Magnetics 2 Question 8
8. Two identical wires $A$ and $B$, each of length $l$, carry the same current $I$. Wire $A$ is bent into a circle of radius $R$ and wire $B$ is bent to form a square of side $a$. If $B _A$ and $B _B$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $\frac{B _A}{B _B}$ is
(2016 Main)
(a) $\frac{\pi^{2}}{8}$
(b) $\frac{\pi^{2}}{16 \sqrt{2}}$
(c) $\frac{\pi^{2}}{16}$
(d) $\frac{\pi^{2}}{8 \sqrt{2}}$
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Answer:
Correct Answer: 8. (d)
Solution:
- $B$ at centre of a circle $=\frac{\mu _0 I}{2 R}$
$B$ at centre of a square
$$ =4 \times \frac{\mu I}{4 \pi \cdot \frac{l}{2}}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]=4 \sqrt{2} \frac{\mu _0 I}{2 \pi l} $$
Now, $\quad R=\frac{L}{2 \pi}$ and $l=\frac{L}{4}$
(as $L=2 \pi R=4 l)$
where, $L=$ length of wire.
$$ \begin{aligned} & \therefore \quad B _A=\frac{\mu _0 I}{2 \cdot \frac{L}{2 \pi}}=\frac{\pi \mu _0 I}{L}=\pi \frac{\mu _0 I}{L} \\ & B _B=4 \sqrt{2} \frac{\mu _0 I}{2 \pi \frac{L}{4}}=\frac{8 \sqrt{2} \mu _0 I}{\pi L}=\frac{8 \sqrt{2}}{\pi} \frac{\mu _0 I}{L} \\ & \therefore \quad \frac{B _A}{B _B}=\frac{\pi^{2}}{8 \sqrt{2}} \end{aligned} $$
