First slide

Magnetic field due to a straight current carrying wire

Question

current r is flowing in a straight conductor of length l. The magnetic induction at a point distant l /4 from its centre will be(fig)

Easy

A

zero
B

$μ_{0} i / (\sqrt{2} l)$
C

$μ_{0} i / (2 πl)$
D

$4 μ_{0} i / (\sqrt{5} πl)$

Solution

See fig. We know that magnetic induction B at a distance r from a long straight conductor carrying a current i is given by
$B = \frac{μ_{0} i}{4 πr} [\sin α + \sin β]$
Here $α = α, β = α and r = l / 4$
$\begin{aligned} ∴ (\sin α + \sin β) = 2 \sin α = 2 \frac{l / 2}{\sqrt{5} l / 4} = \frac{4}{\sqrt{5}} \\ ∴ B = \frac{μ_{0} i}{4 π (l / 4)} \times \frac{4}{\sqrt{5}} = \frac{4 μ_{0} i}{\sqrt{5} πl} \end{aligned}$

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