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Q.

A wire is bent in the form of a quadrant of circle of radius 'a' as shown in the Fig. 8.93. The magnetic field at centre O due to current i is :

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a

zero

b

$\frac{μ_{0}}{4 π} . \frac{πi}{2 a}$

c

$\frac{μ_{0}}{4 π} . \frac{2 πi}{a}$

d

$\frac{μ_{0}}{4 π} . \frac{πi}{a}$

answer is C.

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Detailed Solution

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Magnetic field due to a circular ring at center= $\frac{μ_{o} i}{2 a}$

magnetic field due to quadrant= $\frac{1}{4} \frac{μ_{o} i}{2 a}$

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A wire is bent in the form of a quadrant of circle of radius 'a' as shown in the Fig. 8.93. The magnetic field at centre O due to current i is :