Question

A metal wire $PQ$ slides on parallel metallic rails having separation $0.25m$, each having negligible resistance. There is a $2\Omega$ resistor and $10V$ battery as shown in figure. There is a uniform magnetic field directed into the plane of the paper of magnitude $0.5T$. A force of $0.5N$ to the left is required to keep the wire $PQ$ moving with constant speed is the wire $PQ$ moving?
(Neglect self inductance of the loop)

• A. $8m/s$
• B. $16m/s$
• C. $24m/s$
• D. $32m/s$

Answer 1

The correct option is B $16m/s$

Given: B=0.5T , l=0.25m, F=0.5 N, V=10volt ,R=2ohm

Solution: We know that the force required to move the loop with constant velocity:

$F=Bil=B\times \frac{(V-\epsilon )}{R}\times l$

Where,$\epsilon$=induced emf

$\epsilon =Blv$

$\epsilon =0.5\times 0.25\times v=0.125V$

$0.5N=0.5\times \frac{(10-0.125v)}{2}\times 0.25$

After solving we get that v=16 m/s

Hence the correct answer is B