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Obtain an expression for torque acting on a rectangular current loop. |
Answer» Solution :Let a and b represent breadth and length of a rectangular loop.  Let I be the current in the coil. Let m be the MAGNETIC moment, `tau` be the torque on the coil and B by the magnetic flux linked with the coil. Let B and s represent brush and slip ring. By applying Fleming's LEFT Hand Rule, we note that force on AB and CD remains the same but act opposite to each other. These two and opposite forces constitute couple. DEFLECTING couple or torque is given by the product of magnitude of any one force and perpendicular distance between the forces. Torque, `""tau=Fa` where, `""BIbsintheta=F " on " AB " or " CD.` `""=(BIb)asintheta=IB(ab)sintheta` i.e., `""tau=IBAsintheta ""` where area of the loop = A = ab If `theta=0, sintheta=0` then torque `tau=0` But magnetic moment of the current loop ABCD = IA = m Hence torque on the current loop `=tau=mBsintheta` i.e., `""vec(tau)=vec(m) TIMES vec(B)` Note : As an analogue to torque on electric dipole `""vec(tau)=vec(p)_(e) times vec(E)` where, `p_(e)` - dielectric moment, `vec(E)=` Electric field 
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