A simple pendulum with a bob of mass m and a conducting wire of length L , swings under gravity with an angular amplitude θ . If the horizontal component of the earth's magnetic field perpendicular to the plane of

Correct answer: A. 2BLsin ((θ)/(2)) (gL) ^ 1 / 2 h = L - Lcos θ ⇒ h = L ( 1 - cos θ ) ...(i) using energy conservation Δ KE = Δ PE ⇒ (1)/(2)mv^ 2 = mgh = mgL ( 1 - cosθ ) v= 2gL (2 ^ 2 (θ)/(2)) ⇒ v = 2 gL sin (θ)/(2) Thus, maximum potential difference V _ = BvL = B ×2 gL ((θ)/(2)) L = 2BLsin ((θ)/(2))( gL)1 / 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Electromagnetic Induction

NTA Abhyas 2020 · Physics · Electromagnetic Induction

A simple pendulum with a bob of mass m and a conducting wire of...

A simple pendulum with a bob of mass

m

and a conducting wire of length

L

, swings under gravity with an angular amplitude

\theta

. If the horizontal component of the earth's magnetic field perpendicular to the plane of motion of the pendulum is

B

, then the maximum emf induced across the pendulum is A simple pendulum with a bob of mass m and a conducting wire of length L , swings under gravity with an angular

A simple pendulum with a bob of mass m and a conducting wire of length L , swings under gravity with an angular

Options

A $2BLsin \left(\dfrac{\theta}{2}\right){\left(gL\right)}^{1 / 2}$
B $BLsin \left(\dfrac{θ}{2}\right){\left(2gL\right)}^{1 / 2}$
C $2BLsin \left(\dfrac{θ}{2}\right){\left(gL\right)}^{3 / 2}$
D $BLsin$ \left(\dfrac{θ}{2}\right){ $\left(gL\right)}^{2}$

Correct answer

A. $2BLsin \left(\dfrac{\theta}{2}\right){\left(gL\right)}^{1 / 2}$

Step-by-step solution

h = L - Lcos \theta

⇒ h = L ( 1 - cos \theta )

...(i)
using energy conservation

\Delta KE = \Delta PE

⇒ \dfrac{1}{2}mv^{2} = mgh = mgL ( 1 - cos\theta )

v= 2gL

\left(2\sin^{2} \dfrac{\theta}{2}\right)

⇒ v = 2

\sqrt{gL}

sin

\dfrac{\theta}{2}
Thus, maximum potential difference

{V}_\max = BvL

= B

\times2\sqrt{gL}

\sin \left(\dfrac{θ}{2}\right)

<br />

= 2BLsin \left(\dfrac{θ}{2}\right)\left( gL\right)1 / 2

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