First slide

Magnetic field due to a straight current carrying wire

Question

A square conducting loop of side length L carries a current I. The magnetic field at the centre of the loop is

Moderate

A

independent of L
B

proportional to L²
C

inversely proportional to L
D

linearly proportional to L

Solution

Magnetic field at the centre due to either arm
$\begin{array}{l} B = \frac{μ_{0}}{4 π} \times \frac{I}{(L / 2)} [\sin 45^{0} + \sin 45^{0}] \\ = \frac{μ_{0}}{4 π} \times \frac{2 \sqrt{2} I}{L} \end{array}$
Field at centre due to the four arms of the square
$\begin{array}{l} B = 4 B_{1} = 4 \times \frac{μ_{0}}{4 π} \times \frac{2 \sqrt{2} I}{L} \\ i . e ., B \propto \frac{1}{L} \end{array}$

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