Q12P - Introduction to Electrodynamics

Question

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal:$B\left(t\right)=B_0\cos \left(\omega t\right)\hat{z}$ . A circular loop of wire, of radius $a/2$ and resistance R , is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

Step-by-Step Solution

Answer

The current induced in the loop as function of the time is $\frac{B_0\omega \pi a^2}{4R}\sin \omega t$.

Step 1: Write the given data from the question.

The radius of the solenoid is a .

The magnetic field inside the solenoid,$B\left(t\right)=B_0\cos \left(\omega t\right)\hat{z}$ 

The radius of the circular wire is   and resistance is  R .

Step 2: Calculate the current induced in the loop.

The area of the circular loop is given by,

$$\begin{gathered} \stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}\mathbf{=}\mathit{\pi }{\left(\frac{a}{2}\right)}^{\mathbf{2}}\hat{\mathbf{z}} \\ \stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}\mathbf{=}\frac{\mathbf{\pi }{\mathbf{a}}^{\mathbf{2}}}{\mathbf{4}}\hat{\mathbf{z}} \end{gathered}$$

The magnetic flux through the loop is given by,

$\varphi =\stackrel{\rightharpoonup }{B}.\stackrel{\rightharpoonup }{A}$

                                               

Substitute $\frac{\pi a^2}{4}\hat{z}$ for$\stackrel{\rightharpoonup }{A}$ and$B_0\cos \left(\omega t\right)\hat{z}$ for B into above equation.

$$\begin{gathered} \varphi =B_0\cos \left(\omega t\right)\hat{z}\frac{\pi a^2}{4}\stackrel{ ^}{z} \\ \varphi =B_0\frac{\pi a^2}{4}\cos \left(\omega t\right) \end{gathered}$$

The induced emf in any closed loop is equal to the negative of the rate of change of flux in the circuit.

$\epsilon \left(t\right)=-\frac{d\varphi }{dt}$

Substitute $B_0\frac{\pi a^2}{4}\cos \left(\omega t\right)$ for $\varphi$ into above equation.

$$\begin{gathered} \epsilon \left(t\right)=-\frac{d}{dt}\left(B_0\frac{\pi a^2}{4}\cos \left(\omega t\right)\right) \\ \epsilon \left(t\right)=-B_0\frac{\pi a^2}{4}9\left(-\sin \omega t\right) \omega \\ \epsilon \left(t\right)=\frac{B_0\omega \pi a^2}{4}\sin \omega t \end{gathered}$$

According the ohm’s law, the expression for the current is given by,

$I=\frac{\epsilon \left(t\right)}{R}$

 

Substitute $\frac{B_0\omega \pi a^2}{4}\sin \omega t$ for $\epsilon \left(t\right)$  into above equation.

$$\begin{gathered} I=\frac{\frac{B_0\omega \pi a^2}{4}\sin \omega t}{R} \\ I=\frac{\frac{B_0\omega \pi a^2}{4}\sin \omega t}{4R} \end{gathered}$$

Hence the current induced in the loop as function of the time is $\frac{B_0\omega \pi a^2}{4}\sin \omega t$ .