First slide

Magnetic field due to current element

Question

Equal current i flows in two segments of a circular loop in the direction shown in figure. Radius of the loop is r. The magnitude of magnetic field induction at the centre of the loop is

Moderate

A

zero
B

$\frac{μ_{0} iθ}{3 πr}$
C

$\frac{μ_{0}}{2 π} \frac{i}{r} (π - θ)$
D

$\frac{μ_{0}}{2 π} \frac{i}{r} (2 π - θ)$

Solution

Magnetic field induction at O due to current through ACB is $B_{1} = \frac{μ_{0} iθ}{4 πr}$
It is acting perpendicular to the paper downwards. Magnetic field induction at O due to current through ADB is
$B_{2} = \frac{μ_{0}}{4 π} \frac{i (2 π - θ)}{r}$
It is acting perpendicular to paper upwards.
$∴$ Total magnetic field at O due to current loop is
$\begin{matrix} B = B_{2} - B_{1} = \frac{μ_{0}}{4 π} \frac{i}{r} (2 π - θ) - \frac{μ_{0}}{4 π} \frac{i}{r} θ \\ = \frac{μ_{0}}{2 π} \frac{i}{r} (π - θ) \end{matrix}$

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