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Q.

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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a

2 μT

b

0.5 μT

c

2 μT

d

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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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