The magnitude and direction of magnetic field at point A .

Question

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Answer 1

Question

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2B=μ0I22πri^ (V)

Here, B→2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

B→B=B→1B+B→2B (VI)

Here, B→B is the net magnetic field at point B.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for r in equation (IV).

B→1B=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)(−j^)=4×10−7 T⋅m(50.0 cm(1 m100 cm))(−j^)=(8×10−7 T)(−j^)

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 50.0 cm for r in equation (V).

B→2B=(4π×10−7 T⋅mA)(7.00 A)2π(50.0 cm)i^=14×10−7 T⋅m(50.0 cm(1 m100 cm))i^=(2.80×10−6 T)i^

Substitute −8×10−7j^ T for B→1B and 2.80×10−6i^ T for B→2B in equation (III).

B→B=(8×10−7 T)(−j^)+(2.80×10−6 T)i^=(2.80×10−6 T)i^−(0.80×10−6 T)(j^)

Thus, the magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

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Answer 2

Question

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2B=μ0I22πri^ (V)

Here, B→2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

B→B=B→1B+B→2B (VI)

Here, B→B is the net magnetic field at point B.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for r in equation (IV).

B→1B=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)(−j^)=4×10−7 T⋅m(50.0 cm(1 m100 cm))(−j^)=(8×10−7 T)(−j^)

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 50.0 cm for r in equation (V).

B→2B=(4π×10−7 T⋅mA)(7.00 A)2π(50.0 cm)i^=14×10−7 T⋅m(50.0 cm(1 m100 cm))i^=(2.80×10−6 T)i^

Substitute −8×10−7j^ T for B→1B and 2.80×10−6i^ T for B→2B in equation (III).

B→B=(8×10−7 T)(−j^)+(2.80×10−6 T)i^=(2.80×10−6 T)i^−(0.80×10−6 T)(j^)

Thus, the magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

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Answer 3

Question

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2B=μ0I22πri^ (V)

Here, B→2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

B→B=B→1B+B→2B (VI)

Here, B→B is the net magnetic field at point B.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for r in equation (IV).

B→1B=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)(−j^)=4×10−7 T⋅m(50.0 cm(1 m100 cm))(−j^)=(8×10−7 T)(−j^)

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 50.0 cm for r in equation (V).

B→2B=(4π×10−7 T⋅mA)(7.00 A)2π(50.0 cm)i^=14×10−7 T⋅m(50.0 cm(1 m100 cm))i^=(2.80×10−6 T)i^

Substitute −8×10−7j^ T for B→1B and 2.80×10−6i^ T for B→2B in equation (III).

B→B=(8×10−7 T)(−j^)+(2.80×10−6 T)i^=(2.80×10−6 T)i^−(0.80×10−6 T)(j^)

Thus, the magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

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Answer 4

Question

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2B=μ0I22πri^ (V)

Here, B→2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

B→B=B→1B+B→2B (VI)

Here, B→B is the net magnetic field at point B.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for r in equation (IV).

B→1B=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)(−j^)=4×10−7 T⋅m(50.0 cm(1 m100 cm))(−j^)=(8×10−7 T)(−j^)

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 50.0 cm for r in equation (V).

B→2B=(4π×10−7 T⋅mA)(7.00 A)2π(50.0 cm)i^=14×10−7 T⋅m(50.0 cm(1 m100 cm))i^=(2.80×10−6 T)i^

Substitute −8×10−7j^ T for B→1B and 2.80×10−6i^ T for B→2B in equation (III).

B→B=(8×10−7 T)(−j^)+(2.80×10−6 T)i^=(2.80×10−6 T)i^−(0.80×10−6 T)(j^)

Thus, the magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

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Answer 5

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2B=μ0I22πri^ (V)

Here, B→2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

B→B=B→1B+B→2B (VI)

Here, B→B is the net magnetic field at point B.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for r in equation (IV).

B→1B=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)(−j^)=4×10−7 T⋅m(50.0 cm(1 m100 cm))(−j^)=(8×10−7 T)(−j^)

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 50.0 cm for r in equation (V).

B→2B=(4π×10−7 T⋅mA)(7.00 A)2π(50.0 cm)i^=14×10−7 T⋅m(50.0 cm(1 m100 cm))i^=(2.80×10−6 T)i^

Substitute −8×10−7j^ T for B→1B and 2.80×10−6i^ T for B→2B in equation (III).

B→B=(8×10−7 T)(−j^)+(2.80×10−6 T)i^=(2.80×10−6 T)i^−(0.80×10−6 T)(j^)

Thus, the magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Answer 6

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

Answer 7

Chapter 30, Problem 77PQ

(a)

To determine

The magnitude and direction of magnetic field at point A.

Answer 8

(a)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point A is 3.13×10−6k^ T.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The direction of the magnetic field at the point A is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1A=μ0I12πbk^ (I)

Here, B→1A is the magnetic field due to current I1, μ0 is the permeability of free space, I1 is the current flowing in the wire and b is the distance of point A from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2A=μ0I22πak^ (II)

Here, B→2A is the magnetic field due to current I2 , μ0 is the permeability of free space, I2 is the current flowing in the wire and a is the distance of point A from the wire.

Write the expression for the net magnetic field as.

B→A=B→1A+B→2A (III)

Here, B→A is the net magnetic field at point A.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for b in equation (I).

B→1A=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)k^=4×10−7 T⋅m(50.0 cm(1 m100 cm))k^=(8×10−7 T)k^

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 60.0 cm for a in equation (II).

B→2A=(4π×10−7 T⋅mA)(7.00 A)2π(60.0 cm)k^=14×10−7 T⋅m(60.0 cm(1 m100 cm))k^=(2.33×10−6 T)k^

Substitute 8×10−7k^ T for B→1A and 2.33×10−6k^ T for B→2A in equation (III).

B→A=(8×10−7 T)k^+(2.33×10−6 T)k^=(3.13×10−6 T)k^

Thus, the magnetic field at point A is 3.13×10−6k^ T.

Answer 13

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.

Write the expression for the magnetic field due to wire carrying current I2 as.

B→2B=μ0I22πri^ (V)

Here, B→2B is the magnetic field due to current I2 , μ0 is the permeability of free space and I2 is the current flowing in the wire.

Write the expression for the net magnetic field as.

B→B=B→1B+B→2B (VI)

Here, B→B is the net magnetic field at point B.

Conclusion:

Substitute 4π×10−7 T⋅mA for μ02.00 A for I1 and 50.0 cm for r in equation (IV).

B→1B=(4π×10−7 T⋅mA)(2.00 A)2π(50.0 cm)(−j^)=4×10−7 T⋅m(50.0 cm(1 m100 cm))(−j^)=(8×10−7 T)(−j^)

Substitute 4π×10−7 T⋅mA for μ07.00 A for I2 and 50.0 cm for r in equation (V).

B→2B=(4π×10−7 T⋅mA)(7.00 A)2π(50.0 cm)i^=14×10−7 T⋅m(50.0 cm(1 m100 cm))i^=(2.80×10−6 T)i^

Substitute −8×10−7j^ T for B→1B and 2.80×10−6i^ T for B→2B in equation (III).

B→B=(8×10−7 T)(−j^)+(2.80×10−6 T)i^=(2.80×10−6 T)i^−(0.80×10−6 T)(j^)

Thus, the magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Answer 14

(b)

To determine

The magnitude and direction of magnetic field at point z=−50 cm.

Answer 15

(b)

Expert Solution

Answer to Problem 77PQ

The magnetic field at point z=−50 cm is (2.80×10−6 T)i^−(0.80×10−6 T)(j^).

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Consider the point at z=−50 cm as point B.

The direction of the magnetic field at the point B is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current I1 as.

B→1B=μ0I12πr(−j^) (IV)

Here, B→1B is the magnetic field due to current I1 , μ0 is the permeability of free space, I1 is the current flowing in the wire and r is the distance of point B from the wire.