Question

A metallic square wire loop of side $10cm$ and resistance $1\Omega$ is moved with a constant velocity $v_0$ in a uniform magnetic field of induction $B=2T$ as shown in the figure. The magnetic field is perpendicular to the plane of the loop. The loop is connected to a network of resistors each of value $3\Omega$. The resistance of the lead wires OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of $1mA$ in it ? Give the direction of current in the loop.

• A. $2\times {10}^{-2}$, anticlockwise direction
• B. $4\times {10}^{-2}$, anticlockwise direction
• C. $2\times {10}^{-2}$, clockwise direction
• D. $4\times {10}^{-2}$, clockwise direction

Answer 1

The correct option is C $2\times {10}^{-2}$, clockwise direction

REF.Image

Resistance across QS are forming

balanced Wheatstone bridge

As,

$\frac{R_1}{R_2}=\frac{R_3}{R_4}=\frac{3}{3}=1$

So,

$\frac{1}{Req}=\frac{1}{6}+\frac{1}{6}\Rightarrow Req=3\Omega$

Thus,

Induced emf (Vin) = $BVol$

By ohm's law,

$I(Req+Rw)=Vin$

$\Rightarrow I=\frac{Vin}{(Req+Rw)}=\frac{BVol}{3\Omega +1\Omega }$

As $I=1MA,B=2T$ and $l=\frac{1}{10}m$

$Vo=\frac{\left({10}^{-3}\right)\left(64\right)}{2\times \frac{1}{10}}=2\times {10}^{-2}m/s$

It will flow in clockwise

direction in order to produce magnetic

flux into the page , as it is decreasing

when the coil is moving out the

magnetic field region.

Option - C is correct.