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The magnitude of Magnetic field (in SI units) at the cube of a hexagonal shape coil of side 10cm, 50 turns and carrying current I (ampere) in units of $\frac{\mu _{0}I}{\pi }$ is
Option: 1 $250\sqrt{3}$

Option: 2 $50\sqrt{3}$
Option: 3 $5\sqrt{3}$
Option: 4 $500\sqrt{3}$

Answers (1)

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$B_{A B}=\frac{\mu_{0}}{4 \pi} \times \frac{I}{r}(\sin \theta+\sin \theta)$

$B_{A B}=\frac{\mu_{0}}{4 \pi} \times \frac{2 I}{r} \sin \theta \dots(1)$

Here,

$\theta = 30^o$

and OM=r

$\\ \text{Also from right angled} \ \Delta A M O, we \ have \\ \tan \theta=\frac{A M}{O M}, \\ or, \tan 30^{\circ}=\frac{(a / 2)}{r}$

$\Rightarrow r =\frac{a}{2 \tan 30^{\circ}} = \frac{\sqrt{3} a}{2}$

$r=\frac{\sqrt{3} a}{2} \dots(2)$

From (1) and (2)

$B_ {A B}=\frac{\mu_{0}}{4 \pi} \times \frac{2 I \sin 30^{\circ}}{\sqrt{3} a / 2}=\frac{\mu_{0}}{4 \pi} \times \frac{2 I}{\sqrt{3}a}$

Hence magnetic field due to the whole loop is -

$B_{L oop}=6 \times B_{A B}$

$B_{Loo p}=\frac{\mu_{0} I \sqrt{3}}{\pi a}$

Here,

$n=50, a=0.1 \: m$

So $B_{net }=n*\frac{\mu_{0} I \sqrt{3}}{\pi a}=50*\frac{\mu_{0} I \sqrt{3}}{\pi *0.1}=500\sqrt{3}*\frac{\mu_{0} I }{\pi }$

So, the answer will be -

$500 \sqrt{3}$

Posted by

avinash.dongre

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