First slide

Magnetic field due to a straight current carrying wire

Question

The magnetic field at the centre of coil of n turns, bent in the form of a square of side 2l , carrying current i, is

Moderate

A

$\frac{\sqrt{2} μ_{0} ni}{πl}$
B

$\frac{\sqrt{2} μ_{0} ni}{2 πl}$
C

$\frac{\sqrt{2} μ_{0} ni}{4 πl}$
D

$\frac{2 μ_{0} ni}{πl}$

Solution

Magnetic field due to one side of the square at centre O
$B_{1} = \frac{μ_{0}}{4 π} . \frac{2 isin 45^{0}}{l} \Rightarrow B_{1} = \frac{μ_{0}}{4 π} . \frac{\sqrt{2} i}{l}$
Hence, magnetic field at centre due to all sides
$B = 4 B_{1} = \frac{μ_{0} (\sqrt{2} i)}{π l}$
Magnetic field due to n turns,
$B_{net} = nB = \frac{μ_{0} \sqrt{2} ni}{π l} = \frac{μ_{0} \sqrt{2} ni}{π (l)}$

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