AB is a thin wire of length 2 m carrying a current of 5 A. P is a point on the perpendicular bisector of AB and the distance of P from the wire is 1 m. Then magnetic induction at P is

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2 μT

0.5 μT

2 μT

12 μT

answer is D.

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Detailed Solution

tanα=11⇒α=45o∴B=μ0i2πrsinα=4π x 10-7 x 52π x 1sin45o⇒B=2 × 10-7 × 5 × 12 T=12 μT

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