A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of θ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic

Correct answer: B. mgR θ B^2 l^2 From Faraday's law of electomagnetic induction, aligned e &=(d φ)/(d t)=(d(B A))/(l t)=(d(B l l))/(d t) &=(B d l × l)/(d t)=B V l aligned Also, F=i l B=((B V)/(R))(l^2 B)=(B^2 l^2 V)/(R) At equilibrium m g θ=(B^2 l V)/(R) V= (m g R θ)/(B^2 l^2)

JEE Main 201815 AprilEvening ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Evening Shift
Mode: Online
Subject: Physics
Chapter: Electromagnetic Induction

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Electromagnetic Induction

A copper rod of mass m slides under gravity on two smooth parallel...

A copper rod of mass

\mathrm{m}

slides under gravity on two smooth parallel rails, with separation

l

and set at an angle of

\theta

with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field

\mathrm{B}

normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is: A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of θ

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of θ

Options

A $\frac{\mathrm{mgR} \cos \theta}{\mathrm{B}^2 l^2}$
B $\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$
C $\frac{\mathrm{mgR} \tan \theta}{\mathrm{B}^2 l^2}$
D $\frac{\mathrm{mgR} \cot \theta}{\mathrm{B}^2 l^2}$

Correct answer

B. $\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$

Step-by-step solution

From Faraday's law of electomagnetic induction,

\begin{aligned}e &=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t}\\ &=\frac{B d l \times l}{d t}=B V l \end{aligned}

Also,

F=i l B=\left(\frac{B V}{R}\right)\left(l^2 B\right)=\frac{B^2 l^2 V}{R}

At equilibrium

<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252FIThq93sgx3l5iFAgin_lVatWj22hs8CBE9bN-ro8ftM.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FIThq93sgx3l5iFAgin_lVatWj22hs8CBE9bN-ro8ftM.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FIThq93sgx3l5iFAgin_lVatWj22hs8CBE9bN-ro8ftM.original.fullsize.png?alt=media">

\\ m g \sin

\theta=\frac{B^2 l V}{R} \Rightarrow

V=$\frac{m g R \sin \theta}{B^2 l^2}

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