First slide

Magnetic field due to a straight current carrying wire

Question

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

Moderate

A

$2 μT$
B

$0.5 μT$
C

$\sqrt{2} μT$
D

$\frac{1}{\sqrt{2}} μT$

Solution

$tanα = \frac{1}{1} \Rightarrow α = 45^{o}$
$∴ B = \frac{μ_{0} i}{2 πr} sinα = \frac{4 π x 10^{- 7} x 5}{2 π x 1} \sin 45^{o}$
$\Rightarrow B = 2 \times 10^{- 7} \times 5 \times \frac{1}{\sqrt{2}} T = \frac{1}{\sqrt{2}} μT$

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