SATHEE: Electro Magnetic Induction And Alternating Currents Question 342

Electro Magnetic Induction And Alternating Currents Question 342

Question: A wire cd of length l and mass m is sliding without friction on conducting rails ax and by as shown. The vertical rails are connected to each other with a resistance R between a and b. A uniform magnetic field B is applied perpendicular to the plane abcd such that cd moves with a constant velocity of

Options:

A) $\frac{m g R}{B l}$

B) $\frac{m g R}{B^{2} l^{2}}$

C) $\frac{m g R}{B^{3} l^{3}}$

D) $\frac{m g R}{B^{2} l}$

Show Answer

Answer:

Correct Answer: B

Solution:

Due to magnetic field, wire will experience an upward force

$F = B i l = B (\frac{B v l}{R}) l$

$\Rightarrow F = \frac{B^{2} v l^{2}}{R}$
If wire slides down with constant velocity then
$F = m g \Rightarrow \frac{B^{2} v l^{2}}{R} = m g \Rightarrow v = \frac{m g R}{B^{2} l^{2}}$

Electro Magnetic Induction And Alternating Currents Question 342

Options:

Answer:

Solution:

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Electro Magnetic Induction And Alternating Currents Question 342

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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