First slide

Magnetic field due to a current carrying circular ring

Question

Two identical wires A and 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 B₁ and B₂ are the values of magnetic induction at the centre of the circle and the centre of the square respectively, then
the ratio B₁/B₂ is

Moderate

A

$π^{2} / 8$
B

$π^{2} / 8 \sqrt{2}$
C

$π^{2} / 16$
D

$π^{2} / 16 \sqrt{2}$

Solution

$B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 πI}{R} = \frac{μ_{0}}{4 π} \times \frac{2 πI \times 2 π}{L}$ ….(1)
$\begin{array}{l} (∵ L = 2 πR, for circular loop) \\ B_{2} = \frac{μ_{0}}{4 π} \times \frac{I}{(a / 2)} [\sin 45^{0} + \sin 45^{0}] \times 4 \\ where a = L / 4 \\ ∴ B_{2} = \frac{μ_{0} I}{4 πL} \times 8 \times 4 \times [\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}] \\ = \frac{μ_{0} I}{4 πL} \times \frac{64}{\sqrt{2}} \\ ∴ \frac{B_{1}}{B_{2}} = (\frac{μ_{0}}{4 π}) \frac{4 π^{2} I}{L} / \frac{μ_{0} I}{4 πL} \times \frac{64}{\sqrt{2}} \\ or \frac{B_{1}}{B_{2}} = \frac{π^{2}}{8 \sqrt{2}} \end{array}$

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