Q129 / 160 · WBJEE 2012 · Physics
A straight wire of length $2 \mathrm{m}$ carries a current of $10 \mathrm{A}$. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be
- A$1.5 \mathrm{N}$
- B$3 \mathrm{N}$
- C$3 \sqrt{2} \mathrm{N}$
- Correct answerD$\frac{3}{\sqrt{2}} \mathrm{N}$
Solution
Given $i=10 \mathrm{A}, B=0.15 \mathrm{T}, \theta=45^{\circ}$ and $l=2 \mathrm{m}$
Here,
$$
\begin{aligned}
F &=i L B \sin \theta \\
&=10 \times 2 \times 0.15 ~ x ~ \sin 45^{\circ} \\
&=\frac{3}{\sqrt{2}} \mathrm{N}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.