Question

A uniform magnetic field exists along x-axis, find the flux through a square loop of side a inclined at ${60}^{\circ }$ to $(+\times )$ axis as shown in figure.

• A. $\frac{B{a}^2\sqrt{3}}{2}$
• B. $\frac{{Ba}^2}{2}$
• C. ${Ba}^2$
• D. Zero

Answer 1

The correct option is A $\frac{B{a}^2\sqrt{3}}{2}$

$Solution:$

$\begin{array}{cc}\frac{tg}{\pi }+8\cdot A& =B{a}^2\cos \theta \\ \theta ={30}^{\circ }\text{between}\overrightarrow{B}& \\ & =\frac{\sqrt{3}B{a}^2}{2}\end{array}$

$So,the$ $correct$ $option:A$