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(a) A circular coil of 30 turns and radi...

(a) A circular coil of 30 turns and radius `8.0cm`. Carrying a current of `6.0A` is suspended vertically in a uniform horizontal magnetic field of magnitude `1.0T`. The field lines make an angle of `60^@` with the normal to the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered).

Text Solution

Verified by Experts

Here, `n=30`, `I=6.0A`, `B=1.0T`, `alpha=60^@`, `r=8.0cm=8xx10^-2m`.
`:.` Area of the coil, `A=pir^2=(22)/(7)xx(8xx10^-2)^2=2.01xx10^-2m^2`
(a) Now, `tau=nIBA sin alpha=30xx6.0xx1.0xx(2.01xx10^-2)xxsin60^@`
`=30xx6.0xx2.01xx10^-2xxsqrt3//2=3.133N-m`
(b) Since the torque on the planar loop does not depend upon the shape, in case the area of the loop is the same, the torque will remain unchanged.
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(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60^@ with the normal of the coil. Calculate the magnitude counter torque that must be applied to prevent the coil from turning. (b) Whould your answer change, if the circular coil in (a) were replaced by a planer coil of some irregular shape that encloses the same area? (All other particulars are also unaltered).

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Knowledge Check

  • A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0T. The field lines make an angle of 60^@ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

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