Question

A block of mass 1 g is equilibrium with the help of a current carrying square loop which is partially lying in constant magnetic field (B) as shown. Resistance of the loop is 10 Ω. Find the voltage (V) (in volts) of the battery in the loop.

Answer 1

The correct answer is 10.
ilB = mg
$i=(\frac{mg}{IB})=\frac{(1\times10^{-3}\ kg)\times(10\ m/s^2)}{(0.1\ m)\times(0.1\ T)}$
= 1 × 10–3 × 103
i = 1 A
As resistance of loop = 10 Ω
$i=\frac{V}{R}=1A$
V = (1 × 10) V
= 10 V