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A wire length $\frac{π}{2} m$ , carrying a current of 2A is bent in the form of a circular are which subtends on angle of 45° at its centre. Then magnetic induction at the centre is

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µo / 8 T

µo / 16 T

µo / 2 T

4 µo T

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detailed solution

Correct option is B

l=R.θ⇒R=π/2π/4m=2m ∴Bo=μoi2R.θ2π=μo×22×2×π/42πT ⇒Bo=μo16T

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A wire length π2m, carrying a current of 2A is bent in the form of a circular are which subtends on angle of 45° at its centre. Then magnetic induction at the centre is

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A wire length $\frac{π}{2} m$ , carrying a current of 2A is bent in the form of a circular are which subtends on angle of 45° at its centre. Then magnetic induction at the centre is