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A rectangle loop of size l xx b carrying a steady current I is placed in a uniform magnetic field vecB. Prove that the torque vec(tau) acting on the loop is given by vec(tau) = vecm xx vecB, where vecm is the magnetic moment of the loop. |
Answer» Solution : CONSIDER a rectangular coil PQRS of length l and breadth b, CARRYING current I, placed in a uniform magnetic field B such that a vectro normal to the plane of the coil subtends an angle `theta` from the direction of B. In this situation forces `F_1 and F_2` having magnitude `I b B sin theta` are acting on arms PQ and RS. The forces are equal, opposite nad collinear, hence they cancel out. Again forces `F_3 and F_4` having magnitude `I l B` are acting on each of the two arms QR and SP. These forces too are equal and opposite but these are non-collinear and form a couple whose TORQUE is given as : `tau = (I lB) XX b sin theta = I (lb) B sin theta = I A B sin theta` where `A = I b` = area of the loop. If instead of a single loop, we have a rectangular coil having N turns, then torque `tau = N I A B sin theta` In vector notation `vec(tau) = NI (vecA xx vecB) = (vecm xx vecB) `, where `vecm = N I vecA` = magnetic moment of current carrying coil. |
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