Question

PQ is a uniform rod of length $l$ and mass $m$ carrying current $i$ and is suspended in a uniform magnetic field of induction $\overrightarrow{B}$ acting inward as shown in the figure. The tension in each string is (Given acceleration due to gravity $=g$):

• A. $mg-Bil$
• B. $mg+Bil$
• C. $\frac{mg-Bil}{2}$
• D. $\frac{mg+Bil}{2}$

Answer 1

Answer: (C)

$\frac{mg-Bil}{2}$

Direction of the force will be given by Fleming's left hand rule as shown in the figure.
By this rule, in the given problem, the force on the rod will be in the upward direction.
Assuming tension in each string to be $T$, we can write
$2T+Bil=mg$
$\therefore T=\frac{mg-Bil}{2}$