Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 180.0 $m\Omega$. Assume the field to be uniform.
The magnetic force experienced by the rod when K is closed will be (Assume v = 12 cm s⁻¹):
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $3.75 \times 10^{-3} N$
from ohm's law,
Current, $I=\frac{Blv}{R}=\frac{(0.50)(0.15)(0.12)}{0.180}$
$=0.05A$
and,
$F_{mag}=(0.50)(0.15)(0.05)$
$=0.00375N$
= 3.75 × 10⁻³ N