Q.
A square metallic wire loop of side $0.1\ m$ and resistance of $1\Omega$ is moved with a constant velocity in a magnetic field of $2\ wb/m^{2}$ as shown in figure. The magnetic field is perpendicular to the plane of the loop, loop is connected to a network of resistances. What should be the velocity of loop so as to have steady current of $1mA$ in loop
A) $1\ cm/sec$
B) $2\ cm/sec$
C) $3\ cm/sec$
D) $4\ cm/sec$
Explanation
(b) Equivalent resistance of the given. Wheatstone bridge circuit (balanced) is $3\Omega$ so total resistance in circuit is $R = 3 + 1 = 4\Omega$. The emf induced in the loop $e = Bvl$.
So induced current $i = \frac{e}{R} = \frac{Bvl}{R}$
$\Rightarrow 10^{- 3} = \frac{2 \times v \times ({10 \times 10}^{- 2})}{4} \Rightarrow v = 2cm/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.