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A closed circuit is in the form of a reg...

A closed circuit is in the form of a regular hexagon of side r. If the circuit carries current I, what is the magnetic field induction at the centre of the hexagon?

Text Solution

Verified by Experts

Refer figure, `AG=GB=r//2`
`OG=AG cot 30^@=r//2xxsqrt3`

Magnetic field induction at O due to current in arm AB is
`B_1=(mu_0)/(4pi)xx(I)/((rsqrt3//2))[sin30^@+sin30^@]`
`=(mu_0I)/(2pirsqrt3)`, acting perpendicular to the plane of closed circuit upwards.
The magnetic field induction at O due to currents in arms BC, CD, DE, EF and FA is also equal to `B_1` and acting in same direction. Therefore, total magnetic field induction at O due to current in closed current is
`B=6B_1=6xx(mu_0I)/(2pirsqrt3)=(sqrt3mu_0I)/(pir)`
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