Q 12-06-086JEE MainJEE Main 2023 (10 Apr, Shift 1)Easy
A $1$ m long metal rod $XY$ completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density $0.15$ T. If the resistance of the circuit is $5\ \Omega$, the force needed to move the rod in direction, as indicated, with a constant speed of $4\ \text{m s}^{-1}$ will be ______ $\times10^{-3}$ N.
Numerical value type. Enter your answer.
Answer: 18
$\varepsilon=Blv=0.15\times1\times4=0.6$ V, $I=0.12$ A. The force needed equals the magnetic drag:
$$F=BIl=0.15\times0.12\times1=0.018\ \text{N}=18\times10^{-3}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics