InterviewSolution
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InterviewSolution
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A rectangular coil of sides `8cm` and `6cm` haivng 2000 turns and carrying a current of `200mA` is placed in a uniform magnetic field of `0*2T` directed along the positive x-axis. (a) What is the maximum torque the coil can experience? In which orientation does it experience the maximum Torque? (b) For which orientations of the coil is the torque zero? When is this equilibrium stable and unstable? |
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Answer» We know that a current loop, having n turns, each of area A, carrying current I, when placed in a magnetic field `vecB`, experience a torque whose magnitude is given by `tau=nIAB sin alpha` …(i) where `alpha` is the angle which the normal on the plane of teh current loop makes with the direction of magnetic field , i.e., angle between `vecA` and `vecB`. Here, `n=2000`, `I=200mA=200xx10^-3A`, `A=8xx6sq.cm=48xx10^-4m^2`. `B=0*2T`. (a) Torque acting on the coil will be maximum when `sin alpha=1` or `alpha=90^@` `:.` Maximum Torque, `tau_(max)=nIAB` `=2000xx(200xx10^-3)xx(48xx10^-4)xx0*2` `=0*384N-m` In this situation, the plane of the coil is parallel to the direction of magnetic field i.e., the plane of the coil is in the direction of X-axis. (b) Torque on the coil will be zero, if `sinalpha=0` or `alpha=0^@` or `180^@`. It will be so if plane of the coil is perpendicular to the direction of magnetic field. i.e., the plane of the coil is along Y or Z axis. The coil will be in stable equilibrium when `vecA` is parallel to `vecB` and in unstable equilibrium when `vecA` is antiparallel to `vecB`. |
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