A straight wire AB of mass $40 \mathrm{g}$ and length $50 \mathrm{cm}$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40 \mathrm{T}$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is _____ $\mathrm{A}$. (Take $\left.g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$.

The formula to calculate the magnetic force on the conductor is given by

$F_B= ILB ...\left(1\right)$

The formula to calculate the gravitational force on the conductor is given by

$F_G= mg ...\left(2\right)$

Equate both the equations and simplify to obtain the required current through the conductor.

$\begin{array}{rcl}ILB& =& mg\\ & \Rightarrow & I= \frac{mg}{LB} ...\left(3\right)\end{array}$

Substitute the values of the known parameters into equation (3) to calculate the required current through the conductor.

$\begin{array}{rcl}I& =& \frac{\frac{40}{1000} \mathrm{kg}\times 10 \mathrm{m} {\mathrm{s}}^{-2}}{0.5 \mathrm{m}\times 0.40 \mathrm{T}}\\ & =& 2 \mathrm{A}\end{array}$