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

Question

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle θ. The earth's magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is (A) 2 B L sin (θ/2)(g L)1 / 2 (B) B L sin ((θ/2))(g L) (C) B L sin ((θ/2))(g L)3 / 2 (D) B L sin ((θ/2))(g L)2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3000b0fdcf7698e90e48f922461a54c0-.png" /> ∵ h=L-L cos θ ⇒ h=L(1- cos θ) ∴ v2=2 g h=2 g L(1- cos θ) =2 g L(2 sin 2 θ / 2) ⇒ v=2 √g L sin θ / 2 Thus, maximum potential difference V max =B v L=B × 2 √g L( sin θ / 2) L =2 B L( sin θ / 2)(g L)1 / 2

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

$2 B L sin \frac{θ}{2} (gL)^{1/2}$

$B L sin (\frac{θ}{2}) (gL)$

$B L sin (\frac{θ}{2}) (gL)^{3/2}$

$B L sin (\frac{θ}{2}) (gL)^{2}$

$∵ h = L - L cos θ$
$\Rightarrow h = L (1 - cos θ)$
$∴ v^{2} = 2 g h = 2 gL (1 - cos θ)$
$= 2 gL (2 sin^{2} θ /2)$
$\Rightarrow v = 2 gL sin θ /2$
Thus, maximum potential difference
$V_{m a x} = B vL = B \times 2 gL (sin θ /2) L$
$= 2 B L (sin θ /2) (gL)^{1/2}$

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

2BLsin2θ​(gL)1/2

BLsin(2θ​)(gL)

BLsin(2θ​)(gL)3/2

BLsin(2θ​)(gL)2