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A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil?A. 1.6 N mB. 1.2 N mC. 1.4 N mD. 1.8 N m |
Answer» Correct Answer - A `tau=NIABsin theta` Here, `N=25, I=10A, B=0.9T, theta=30^(@)` `A=a^(2)=12xx10^(-2)xx12xx10^(-2)=144xx10^(-4)m^(2)` `therefore" "tau=25xx10xx144xx10^(-4)xx0.9xxsin 30^(@)="1.6 N m"` |
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