Question

The magnetic force per unit length on a wire carrying a current of $10\, A$ and making an angle of $45^{\circ}$ with the direction of a uniform magnetic field of $0.20\, T$ is

• A. $2 \sqrt{2} \, N \,m^{-1}$
• B. $\frac{2}{\sqrt{2}} N \, m^{-1}$
• C. $\frac{\sqrt{2}}{2}N \,m^{-1}$
• D. $4 \sqrt{2}N \,m^{-1}$

Answer 1

Answer: (B)

$I=10 \,A$, $\theta=45^{\circ}$, $B=0.2\,T$
$\therefore F=IlB \, sin \,\theta $
$\therefore \frac{F}{l}=IB \, sin \,45^{\circ} =10\times0.2 \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}} N \, m^{-1}$