First slide

Magnetic flux and induced emf

Question

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle $2 θ$ .The earth's magnetic field component in the direction perpendicular to swing is, B. Maximum potential difference induced across the pendulum is

Moderate

A

$2 B L \sin (\frac{θ}{2}) (g L)^{1 / 2}$
B

$B L \sin (\frac{θ}{2}) (g L)$
C

$B L \sin (\frac{θ}{2}) (g L)^{3 / 2}$
D

$B L \sin (\frac{θ}{2}) (g L)^{2}$

Solution

From figure,
$\Rightarrow h = L (1 - \cos θ)$
Maximum velocity at equilibrium is given by
$\begin{array}{l} ∴ v^{2} = 2 g h = 2 g L (1 - \cos θ) = 2 g L (2 \sin^{2} \frac{θ}{2}) \end{array}$
$\Rightarrow v = 2 \sqrt{g L} \sin \frac{θ}{2}$
Thus, maximum potential difference
$V_{max} = B v L = B \times 2 \sqrt{g L} \sin \frac{θ}{2} L$
$= 2 B L \sin \frac{θ}{2} (g L)^{1 / 2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App