Chapter 27, Problem 27P

(a)

To determine

The magnetic field at the point $y=-3\text{cm}$

Answer to Problem 27P

The magnetic field at the point $y=-3\text{cm}$ is $-\left(0.\text{18}\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .

Explanation of Solution

Given:

The current in the wire at $y=-\text{6}\text{cm}$ and $y=+\text{6}\text{cm}$ is in the $-x$ direction and $+x$ direction respectively.

The magnitude of current is $I=20\text{A}$

Formula used:

The expression for the magnetic field in the wire is given by,

  $B=\frac{{\mu }_{o}I}{2\pi R}$

Calculation:

The magnetic field at point $y=-3\text{cm}$ that is at the distance of $3\text{cm}$ is calculated as,

  $\begin{array}{l}{B}_1=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(6\times {10}^{-2}\text{m}-3\times {10}^{-2}\text{m}\right)}\\ =13.34\times {10}^{-5}\text{T }\end{array}$

The sign is given by using the right hand thumb rule as,

  $B_1=-13.34\times {10}^{-5}\text{T <mover accent="true"><mo>k</mo><mo>^</mo></mover>}$

Now at $y=-3\text{cm}$ that is at the distance of $9\text{cm}$

  $\begin{array}{l}{B}_2=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\left(20\text{A}\right)}{2\pi \left(9\times {10}^{-2}\text{m}\right)}\\ =4.44\times {10}^{-5}\text{T }\end{array}$

The sign of magnetic field is given by using the right hand thumb rule,

  $B_2=-4.44\times {10}^{-5}\text{T<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$

The net magnetic field is given as:

  $\begin{array}{c}B=\left(-4.44\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>+(}-\text{13}\text{.34}\times {10}^{-5}\text{T)<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ =-17.78\times {10}^{-5}\text{T}\times \left(\frac{\text{1000}\text{mT}}{\text{1}\text{T}}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ \approx -\left(0.\text{18}\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

Conclusion:

Therefore, the magnetic field at a point is $-\left(0.\text{18}\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .

(b)

To determine

The magnetic field at the point $y=0$

Answer to Problem 27P

The magnetic field at the point $y=0$ is $-\left(0.13\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .

Explanation of Solution

Calculation:

The magnetic field at point $y=0$ is calculated as,

  $\begin{array}{l}{B}_1=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(6\times {10}^{-2}\text{m}\right)}\\ =6.67\times {10}^{-5}\text{T }\end{array}$

The sign of magnetic field is given by using the right hand thumb rule,

  $B_1=-\left(6.67\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover><mo> </mo>}$

Now at $y=0$ that is $y=6\text{cm}$

  $\begin{array}{l}{B}_2=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(6\times {10}^{-2}\text{m}\right)}\\ =6.67\times {10}^{-5}\text{T }\end{array}$

Similarly, the sign of magnetic field is given by using the right hand thumb rule,

  $B_2=-\left(6.67\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover><mo> </mo>}$

The net magnetic field is given as:

  $\begin{array}{c}B=\text{(}-6.67\times {10}^{-5}\text{T)<mover accent="true"><mo>k</mo><mo>^</mo></mover>+(}-6.67\times {10}^{-5}\text{T)<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ =-13.34\times {10}^{-5}\text{T}\times \left(\frac{\text{1000}\text{mT}}{\text{1}\text{T}}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ =-\left(0.13\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

Conclusion:

Therefore, the magnetic field at a point is $-\left(0.13\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .

(c)

To determine

The magnetic field at the point $y=+3\text{cm}$

Answer to Problem 27P

The magnetic field at the point is $\left(-0.\text{18}\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$

Explanation of Solution

Calculation:

The magnetic field at point $y=+3$ that is at the distance of $9\text{cm}$ is calculated as,

  $\begin{array}{l}{B}_1=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(9\times {10}^{-2}\text{m}\right)}\\ =\left(-4.44\times {10}^{-5}\text{T}\right)\text{ <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

Now At $y=+3$ that is at the distance of $3\text{cm}$

  $\begin{array}{c}{B}_2=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(3\times {10}^{-2}\text{m}\right)}\\ =\left(-13.34\times {10}^{-5}\text{T}\right)\text{ <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

The net magnetic field is given as:

  $\begin{array}{c}B=\left(-4.44\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>+(}-\text{13}\text{.34}\times {10}^{-5}\text{T)<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ =-17.78\times {10}^{-5}\text{T}\times \left(\frac{\text{1000}\text{mT}}{\text{1}\text{T}}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\\ \approx \left(-0.\text{18}\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

Conclusion:

The magnetic field at the point is $\left(-0.\text{18}\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .

(d)

To determine

The magnetic field at the point $y=+9\text{cm}$

Answer to Problem 27P

The magnetic field at the point is $\left(0.11\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$.

Explanation of Solution

Calculation:

The magnetic field at point $y=+9$ that is at the distance of $15\text{cm}$ is calculated as,

  $\begin{array}{c}{B}_1=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(15\times {10}^{-2}\text{m}\right)}\\ =-\left(2.67\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover><mo> </mo>}\end{array}$

Now At $y=+9$ that is at the distance of $-3\text{cm}$

  $\begin{array}{l}{B}_2=\frac{{\mu }_{o}I}{2\pi R}\\ =\frac{\left(4\pi \times {10}^{-7}\right)\times \left(20\text{A}\right)}{2\pi \left(-3\times {10}^{-2}\text{m}\right)}\\ =\left(13.4\times {10}^{-5}\text{T}\right)\text{ <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

The net magnetic field is given as:

  $\begin{array}{c}B=-\left(2.67\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover><mo> ;+</mo>}\left(13.34\times {10}^{-5}\text{T}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover><mo> </mo>}\\ =\left(10.67\times {10}^{-5}\text{T}\right)\times \left(\frac{\text{1000}\text{mT}}{\text{1}\text{T}}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover><mo> </mo>}\\ \approx \left(0.11\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}\end{array}$

Conclusion:

The magnetic field at the point is $\left(11\text{mT}\right)\text{<mover accent="true"><mo>k</mo><mo>^</mo></mover>}$ .