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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 <img alt="\frac{\mu _{0}I}{\pi }" src="http://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7B%5

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)

best_answer

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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