Question

A plane spiral with a great number N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral’s plane the outside radius of the spiral’s turns is equal to a and inner radius is zero. The magnetic induction various with time as $B = {B_0}\sin \omega t$ where ${B_0}$ and $\omega$ are constants find the amplitude of emf induced the spiral.
(A) $\dfrac{1}{3}\pi {a^2}N\omega {B_0}$
(B) $\dfrac{1}{2}\pi {a^2}N\omega {B_0}$
(C) $\dfrac{1}{4}\pi {a^2}N\omega {B_0}$
(D) None of these

Answer 1

Here,we are going to apply the concept of induced emf and Lenz’s law and in the given problem magnetic field is given in terms of time. So, first calculate the emf for one turn and then multiply it with N number of turns to get the required answer.

Formula used:
$e = \dfrac{{ - d{\phi _B}}}{{dt}}$
Where,
${\phi _B} =$ Magnetic flux $(\overrightarrow B .\overrightarrow A )$
B $=$ magnetic field, A $=$ Area

Complete step by step answer:
Given that the plane spiral shape is made up of concentric loops, having different radii from 0 to a.
We know that induced emf due to loop is
$e = \dfrac{{ - d{\phi _B}}}{{dt}}$
${\phi _B} = \overrightarrow B .\overrightarrow A$
So, $e = \dfrac{{ - d(\overrightarrow B .\overrightarrow A )}}{{dt}}$
Given that $B = {B_0}\sin \omega t$
So, $e = - A\dfrac{{dB}}{{dt}}$
Where A $=$ area i.e., $\pi {r^2}$
$e = - \pi {r^2}\dfrac{{d({B_0}\sin \omega t)}}{{dt}}$
$\Rightarrow e = - {B_0}\pi {r^2}\omega \cos \omega t$ …...(1)
So, the total induced emf is
$e = - \int\limits_0^a {(\pi {r^2}{B_0}\omega \cos \omega t)dN}$ …..(2)
Where $\pi {r^2}\omega \cos \omega t$ is the contribution of one turn of radius r.
dN $=$ Number of turns in the interval r to $r + dr$
$\Rightarrow dN = \left( {\dfrac{N}{a}} \right)dr$ …..(3)
From equation 2 and 3
$\Rightarrow\varepsilon = - \int\limits_0^a {(\pi {r^2}{B_0}\omega \cos \omega t)} \dfrac{N}{a}dr$
$\Rightarrow\varepsilon = - \pi {B_0}\omega \dfrac{N}{a}\cos \omega t\int\limits_0^a {{r^2}dr}$
$\Rightarrow\varepsilon = - \pi {B_0}\omega \dfrac{N}{a}\cos \omega t\left( {\dfrac{{{r^3}}}{3}} \right)_0^a$
$\Rightarrow\varepsilon = \dfrac{{ - \pi {B_0}\omega N\cos \omega t}}{a}\left( {\dfrac{{{a^3}}}{3} - 0} \right)$
$\therefore\varepsilon = - \dfrac{1}{3}\pi {a^2}{B_0}N\omega \cos \omega t$
Hence the amplitude of emf induced in spiral is $\dfrac{1}{3}\pi {a^2}{B_0}N\omega$
So, option A is the correct answer.

Note: In problems of induced emf students may get confused between emf for one turn and for complete spiral. So, always remember to multiply one turn emf with the total number of turns.