Question

A current of $10$ $A$ is flowing in a wire of length $1.5m$. A force of $15$ $N$ acts on it in magnetic field of $2T$. The angle between the magnetic field and the direction of current is:

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (A)

${30}^o$

Force on a current carrying wire of length $l$ and current $i$, placed in uniform magnetic field $B$ is given by

$F=i(\overrightarrow{l}\times \overrightarrow{B})$
$F=IlBsin\theta$ or $sin\theta =\frac{F}{IlB}$

$sin\theta =\frac{15}{10\times 1.5\times 2}=\frac{1}{2}$ or $\theta ={30}^o$