Question

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) B cos(ot)ż. 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. 6.

Answer 1

Answer:

The current induced in the loop is $\dfrac{\pi a^2 B\omega}{4R}\ \sin(\omega t)$.

Explanation:

Given that,

Magnetic field $B(t)=B\cos(\omega t)\ z$

Radius $r =\dfrac{a}{2}$

Resistance =R

We need to calculate the area of the loop

Using formula of area

$A = \pi r^2$

Put the value of r in to the formula

$A =\pi\times(\dfrac{a}{2})^2$

$A=\dfrac{\pi a^2}{4}$

We need to calculate the flux

Using formula of flux

$\phi=BA$

$\phi=B\cos(\omega t)\dfrac{\pi a^2}{2}$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{-\pi a^2B}{4}(-\omega\sin\omega t)$

$\epsilon=\dfrac{\pi a^2B\omega\sin(\omega t)}{4}$

We need to calculate the current

Using formula of current

$I(t)=\dfrac{\epsilon}{R}$

$I(t)=\dfrac{\dfrac{\pi a^2B\omega\sin(\omega t)}{4}}{R}$

$I(t)=\dfrac{\pi a^2 B\omega}{4R}\ \sin(\omega t)$

Hence, The current induced in the loop is $\dfrac{\pi a^2 B\omega}{4R}\ \sin(\omega t)$.