First slide

Magnetic field due to a current carrying circular ring

Question

The loop shown in figure in carrying a current I the magnetic induction at the common centre O of the semi circles is B.
Now the wire of the loop is bent to form a circular ring which is carrying the same current I. Then magnetic induction at the centre of the ring will be

Moderate

A

$8 π B / 3 (3 π + 2)$
B

$4 π B / (π + 2)$
C

$8 B / (π + 4)$
D

$2 B / (π + 8)$

Solution

$B = \frac{μ_{0} I}{4 R} + \frac{μ_{0} I}{4 (\frac{R}{2})} = \frac{3 μ_{0} I}{4 R}$
If radius of the circular ring be $R^{1}$ , then
$2 π R^{1} = π R + π (\frac{R}{2}) + R$
$\Rightarrow R^{1} = \frac{3 π R}{2 \times 2 π} + \frac{R}{2 π} = (\frac{3 π + 2}{4 π}) R$
$∴ B^{1} = \frac{μ_{0} I}{2 R^{1}} = \frac{μ_{0} I}{R (3 π + 2) / 2 π} = \frac{μ_{0} I \times 2 π}{R (3 π + 2)}$
$\Rightarrow B^{1} = \frac{2 π}{(3 π + 2)} \times \frac{4 B}{3} = \frac{8 π B}{3 (3 π + 2)}$

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