Two identical wires A a n d B have the same length L and carry the same current | TestHub

Two identical wires $A a n d B$ have the same length $L$ and 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 $B_{1} a n d B_{2}$ are the value of magnetic induction at the centre of the circle and at the centre of the square respectively, then the ratio $\frac{B_{1}}{B_{2}}$ is

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Answer:

Solution:

As, $B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 π I}{R}$

$= \frac{μ_{0}}{4 π} \times \frac{2 π I \times 2 π}{L}$ ... (i)

$[∵ L = 2 π R f o r c i r c u l a r l o o p]$

$B_{2} = \frac{μ_{0}}{4 π} \times \frac{I}{(\frac{a}{2})} [s i n 45^{o} + s i n 45^{o}] \times 4$

Where $a = \frac{L}{4}$

$∴ B_{2} = \frac{μ_{0} I}{4 π L} \times 8 \times 4 \times [\frac{I}{\sqrt{2}} + \frac{I}{\sqrt{2}}]$

$= \frac{μ_{0} I}{4 π L} \times \frac{64}{\sqrt{2}}$ ..... (ii)

Hence, $\frac{B_{1}}{B_{2}} = (\frac{μ_{0}}{4 π}) \times \frac{\frac{4 π^{2} I}{L}}{\frac{μ_{0} I}{4 π L} \times \frac{64}{\sqrt{2}}}$

Or $\frac{B_{1}}{B_{2}} = \frac{π^{2}}{8 \sqrt{2}}$

Physics - EMI/AC Question with Solution | TestHub

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