Question

If a square loop made with a wire of uniform cross section current $I$ enters from point $A$ and leaves from point $B$. The magnetic field strength $B$ at the centre of the square is

• A. $zero$
• B. $\frac{{\mu }_{0}I2\sqrt{2}}{4\pi a}$
• C. $\frac{4\sqrt{2}{\mu }_{0}I}{4\pi a}$
• D. $\frac{2\sqrt{2}{\mu }_{0}I}{4a}$

Answer 1

The correct option is A $zero$

As the material of the loop is same, resistance through the path with 3 sides will be 3 times the resistance through the path with one side. So current will get divided in the ratio shown in the figure.
Also, ${\theta }_1$ and ${\theta }_2$ mentioned in the formula will be equal to ${45}^{\circ }$ as the loop is square loop. ( can be deduced from the figure)
The magnetic field due to the shorter path will be $\frac{{\mu }_0.0.75I}{4\pi .\frac{a}{\sqrt{2}}}$
The magnetic field due to the longer path will be $3\times \frac{{\mu }_0.0.25I}{4\pi .\frac{a}{\sqrt{2}}}=\frac{{\mu }_0.0.75I}{4\pi .\frac{a}{\sqrt{2}}}$
By Ampere's right hand grip rule, we can conclude that the magnetic field due to both these paths is opposite in direction. So, the fields will cancel out and the resultant field at the center will be zero.