First slide

Magnetic field due to current element

Question

A current I flows through a wire of finite length L. The magnetic field at a distance d from the wire on its perpendicular bisector,

Moderate

A

$\frac{μ_{o} I}{2 π d} \frac{L}{\sqrt{L^{2} + 4 d^{2}}}$
B

$\frac{μ_{o} I}{4 π d} \frac{L}{\sqrt{L^{2} + 4 d^{2}}}$
C

$\frac{μ_{o} I}{π d} \frac{2 L}{\sqrt{L^{2} + 4 d^{2}}}$
D

Zero

Solution

$P o i n t i s o n p e r p e n d i c u l a r b i s e c t o r, s o, α = β F i e l d d u e t o f i n i t e w i r e = B = \frac{μ_{o} I}{4 π d} [S i n α + S i n β] = \frac{μ_{o} I}{4 π d} [2 S i n α] = \frac{μ_{o} I}{2 π d} S i n α = \frac{μ_{o} I}{2 π d} \frac{L}{\sqrt{L^{2} + 4 d^{2}}}$

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