The magnitude and the direction of the magnetic field at the origin.

Question

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Answer 1

Question

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

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Answer 2

Question

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

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Answer 3

Question

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

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Answer 4

Question

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

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Answer 5

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

Answer 6

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Answer 7

Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

Answer 8

(a)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$ (1)

Here, $μ0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1−B2)$ (2)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1= 4π×10−7(10 A)2π(0.1 m)=2×10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B1$ for $B$ , $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $d$ in equation (1).

$B1=(4π×10−7)(5 A)2π(0.1 m)=10−5 T$

Substitute $2×10−5 T$ for $B1$ and $10−5 T$ for $B2$ in equation (2).

$B=(2×10−5 T−10−5 T)=10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $10−5 T$ and the magnitude is out of page.

Answer 13

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Answer 14

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

Answer 15

(b)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$ -axis due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=−(B1+B2)$ (3)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(x−a)$ for $d$ in equation (1).

$B1=μ0I2π(x−a)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m−0.1 m)=10−5 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m+0.1 m)=6.67×10−6 T$

Substitute $10−5 T$ for $B1$ and $6.67×10−6 T$ for $B2$ in equation (3).

$B=−(10−5 T+6.67×10−6 T)=−1.67×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67×10−5 T$ and the magnitude is in page.

Answer 20

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Answer 21

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

Answer 22

(c)

Expert Solution

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$ -axis due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (4)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B1$ for $B$ and $(a+x)$ for $d$ in equation (1).

$B1=μ0I2π(a+x)$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute $(4π×10−7)$ for $μ0$ , $5 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B1=(4π×10−7)(5 A)2π(0.2 m+0.1 m)=3.33×10−6 T$

The magnetic field due to the second wire is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(x−a)$ for $d$ in equation (1).

$B2=μ0I′2π(x−a)$

Substitute $(4π×10−7)$ for $μ0$ , $10 A$ for $I$ and $0.1 m$ for $a$ and $0.2 m$ for $x$ in the above equation.

$B2=(4π×10−7)(10 A)2π(0.2 m−0.1 m)=2×10−5 T$

Substitute $3.33×10−6 T$ for $B1$ and $2×10−5 T$ for $B2$ in equation (4).

$B=(3.33×10−6 T+2×10−5 T)=2.33×10−5 T$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33×10−5 T$ and the magnitude is out of the page.

Answer 27

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

Answer 28

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .

Answer 29

(d)

Expert Solution

Answer to Problem 39E

The distance is $0.0333 m$ at which the magnetic field is zero

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

$B=μ0I2πd$

Write the expression for the net magnetic field due to two parallel conducting wire.

$B=(B1+B2)$ (5)

Here, $B1$ is the field due to first wire and $B2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$ -axis is calculated below.

Substitute $B1$ for $B$ and $(a−x)$ for $d$ in equation (1).

$B1=μ0I2π(a−x)$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$ -axis is calculated below.

Substitute $B2$ for $B$ , $I′$ for $I$ and $(a+x)$ for $d$ in equation (1).

$B2=μ0I′2π(a+x)$

Substitute $μ0I2π(a−x)$ for $B1$ and $μ0I′2π(a+x)$ for $B2$ in equation (5).

$B=μ0I2π(a−x)−μ0I′2π(a+x)$

Substitute $0$ for $B$ and rearrange the above equation.

$0=μ0I2π(a−x)−μ0I′2π(a+x)I(a−x)=I′(a+x)$

Substitute $10 A$ for $I$ , $5 A$ for $I′$ and $0.1 m$ for $a$ in the above equation.

$10 A(0.1 m−x)=5 A(0.1 m+x)2(x+0.1 m)=(0.1 m−x)$

Simplify the above equation.

$x≈0.0333 m$

Thus, the distance is $0.0333 m$ at which the magnetic field is zero.

Answer 34

Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ

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Answer to Problem 39E

Explanation of Solution

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Answer to Problem 39E

Explanation of Solution

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Answer to Problem 39E

Explanation of Solution

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Answer to Problem 39E

Explanation of Solution

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Chapter 19 Solutions

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Answer to Problem 39E

Explanation of Solution

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Answer to Problem 39E

Explanation of Solution

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Answer to Problem 39E

Explanation of Solution

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Answer to Problem 39E

Explanation of Solution

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Chapter 19 Solutions

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

The magnitude and the direction of the magnetic field at any point in the positive $x$ -axis.

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$ .