Question

A horizontal rod $0.2\text{m}$ long is mounted on a balance and carries a current. At the location of the rod, a uniform horizontal magnetic field has a magnitude $0.067\text{T}$ and direction perpendicular to the rod. The magnetic force on the rod is measured by a balance and found to be $0.13\text{N}$. What is the current ?

• A. $0.13\text{A}$
• B. $9.7\text{A}$
• C. $6.2\text{A}$
• D. $4.5\text{A}$

Answer 1

The correct option is B $9.7\text{A}$

Here, the direction of $\overrightarrow{B}$ is perpendicular to the plane of rod. So angle between length vector and magnetic field $\overrightarrow{B}$ is ${90}^{\circ }$.

$\therefore \theta ={90}^{\circ }$

Applying formula of magnetic force on current carrying wire,

$\overrightarrow{F}=i(\overrightarrow{L}\times \overrightarrow{B})$

$F=iLB\sin {90}^{\circ }$

$i=\frac{F}{LB\sin {90}^{\circ }}=\frac{0.13}{0.2\times 0.067}$

$\therefore i=9.7\text{A}$

Therefore, option $\left(B\right)$ is correct.