First slide

Magnetic field due to current element

Question

Figure shows a wire PQRS carrying a current I. Portions PQ and RS are straight and QR in a circular arc of radius r subtending an angle $α$ at the centre C. The magnitude of magnetic field due to PQRS at centre C is

Moderate

A

$\frac{μ_{0} I}{r} (1 + \frac{α}{2 π})$
B

$\frac{μ_{0} I}{r} α$
C

$\frac{μ_{0} Iα}{2 πr}$
D

$\frac{μ_{0} Iα}{4 πr}$

Solution

The straight portions PQ and RS do not produce magnetic field at C. Only the curved portion QR produces a magnetic field at C. Refer to Fig.
Since the current dl is perpendicular to position vector r, the magnetic field at C by Biot-Savart law is
$\begin{array}{l} dB = \frac{μ_{0}}{4 π} \frac{Idl \sin 90}{r^{2}} \\ = \frac{μ_{0}}{4 π} \frac{I \times (rdθ)}{r^{2}} (∵ dl = rd \underline{θ}) \\ = \frac{μ_{0} I}{4 πr} dθ \end{array}$
$B = \int_{0}^{α} 1 d B = \frac{μ_{0} I}{4 πr} \int_{0}^{α} dθ = \frac{μ_{0} Iα}{4 πr}$
So the correct choice is (4).

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