Chapter 32, Problem 13PQ

A square conducting loop with side length a = 1.25 cm is placed at the center of a solenoid 40.0 cm long with a current of 4.30 A flowing through its 420 turns, and it is aligned so that the plane of the loop is perpendicular to the long axis of the solenoid. The radius of the solenoid is 5.00 cm. a. What is the magnetic flux through the loop? b. What is the magnitude of the average emf induced in the loop if the current in the solenoid is increased from 4.30 A to 10.0 A in 1.75 s?

To determine

The magnetic flux through the loop.

Answer to Problem 13PQ

The magnetic flux through the loop is $8.87\times {10}^{-7}\text{Wb}$.

Explanation of Solution

Write the expression for the magnetic field of solenoid.

  $\begin{array}{c}B={\mu }_{0}nI\\ ={\mu }_0\left(\frac{N}{l}\right)I\end{array}$                                                                                                           (I)

Here, $B$ is the magnetic field inside the solenoid, $N$ is the total number of turns, $l$ is the length of the solenoid, and $I$ is current in the solenoid.

Write the expression for the magnetic flux through the loop.

  $\begin{array}{c}{\varphi }_B=BA\cos \phi \\ =B{a}^2\cos \phi \end{array}$                                                                                                            (II)

Here, $A$$\left(=a^2\right)$ is area of the loop and $\phi$ is the angle between the magnetic field and area vector.

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{H/m}$ for ${\mu }_0$, $420$ for $N$, $40.0\text{cm}$ for $l$ and $4.30\text{A}$ for $I$ in equation (I) to find $B$.

  $\begin{array}{c}B=\left(4\pi \times {10}^{-7}\text{H/m}\right)\times \left(\frac{420}{40.0\text{cm}\times \frac{{10}^{-2}{\text{m}}^{\text{2}}}{1\text{cm}}}\right)\times \left(4.30\text{A}\right)\\ =5.67\text{mT}\end{array}$

Thus, the magnetic field inside the solenoid is $5.67\text{mT}$.

Substitute $5.67\text{mT}$ for $B$, $1.25\text{cm}$ for $a$ and $0°$ for $\phi$ in equation (II) to find ${\varphi }_B$.

  $\begin{array}{c}{\varphi }_B=\left(5.67\text{mT}\right)\times {\left(1.25\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\times \cos 0°\\ =8.87\times {10}^{-7}\text{Wb}\end{array}$

Therefore, the magnetic flux through the loop is $8.87\times {10}^{-7}\text{Wb}$.

To determine

The magnitude of the average emf induced in the loop.

Answer to Problem 13PQ

The magnitude of the average emf induced in the loop is $6.72\times {10}^{-7}\text{V}$

Explanation of Solution

Write the expression for Faraday’s law of induction.

  $\left|\epsilon \right|=\left|\frac{d{\varphi }_B^{}}{dt}\right|$

Here, $\epsilon$ is the emf induced in the loop and ${\varphi }_{\text{B}}$ is the magnetic flux associated with the loop.

Substitute the value of equation (II) in the above equation to find $\left|\epsilon \right|$.

  $\left|\epsilon \right|=\left|\frac{d\left(B{a}^2\cos \phi \right)}{dt}\right|$

Conclusion:

Substitute the value of equation (I) in the above equation to find $\left|\epsilon \right|$.

  $\begin{array}{c}\left|\epsilon \right|=\left|\frac{d\left({\mu }_0\left(\frac{N}{l}\right)I{a}^2\cos \phi \right)}{dt}\right|\\ ={\mu }_0\left(\frac{N}{l}\right)a^2\cos \phi \left|\frac{I_2-I_1}{dt}\right|\end{array}$

Substitute $4\pi \times {10}^{-7}\text{H/m}$ for ${\mu }_0$, $420$ for $N$, $40.0\text{cm}$ for $l$, $1.25\text{cm}$ for $a$, $0°$ for $\phi$, $10.0\text{A}$ for $I_2$, $4.30\text{A}$ for $I_1$ and $1.75\text{s}$ for $\Delta t$  in equation (V) to find $\epsilon$.

  $\begin{array}{c}\left|\epsilon \right|=\left(4\pi \times {10}^{-7}\text{H/m}\right)\frac{420}{\left(40.0\text{cm}\times \frac{{10}^{-2}{\text{m}}^{\text{2}}}{1\text{cm}}\right)}{\left(1.25\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\cos 0°\times \left|\left(\frac{10.0\text{A}-4.30\text{A}}{1.75\text{s}}\right)\right|\\ =6.72\times {10}^{-7}\text{V}\end{array}$

Thus, the magnitude of the average emf induced in the loop is $6.72\times {10}^{-7}\text{V}$