A square loop ABCD of side a carrying current I is folded about an axis passing through its centre (y-axis). The two halves are inclined at angle 300 with xy plane as shown. -Find the magnetic fieid at origin. The origin coincides with the centre of loop.

Bnet = 2Bhalf cos θWhere Bhalf = 12 4μ04πa Ia/2 cos450 + cos 450=μ0I2πaSo B=2 μ0I2πa 32 = 6μ0I πa

A square loop ABCD of side a carrying current I is folded about an axis passing through its centre (y-axis). The two halves are inclined at angle 300 with xy plane as shown. -Find the magnetic fieid at origin. The origin coincides with the centre of loop.

Bnet = 2Bhalf cos θWhere Bhalf = 12 4μ04πa Ia/2 cos450 + cos 450=μ0I2πaSo B=2 μ0I2πa 32 = 6μ0I πa

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A square loop ABCD of side a carrying current I is folded about an axis passing through its centre (y-axis). The two halves are inclined at angle 30⁰ with xy plane as shown. -Find the magnetic fieid at origin. The origin coincides with the centre of loop.

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$\frac{μ_{0} I \sqrt{3}}{\sqrt{2} πa}$

$\frac{\sqrt{6} μ_{0} I}{πa}$

$\frac{μ_{0} I \sqrt{3}}{2 \sqrt{2} πa}$

$\frac{3 μ_{0} I}{2 \sqrt{2} πa}$

answer is D.

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Detailed Solution

$B_{net} = 2 (B_{half}) \cos θ$

$\begin{array}{l} Where B_{half} = \frac{1}{2} [\frac{4 μ_{0}}{4 πa} \frac{I}{(a / 2)}] (\cos 45^{0} + \cos 45^{0}) \\ = \frac{μ_{0} I \sqrt{2}}{πa} \\ So B = 2 (\frac{μ_{0} I \sqrt{2}}{πa}) \frac{\sqrt{3}}{2} = \frac{\sqrt{6} μ_{0} I}{πa} \end{array}$