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Compute the torque experimenced by a magnetic needle in a uniform magnetic field. |
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Answer» Solution :Torque Acting on a Bar Magnet in Uniform Magnetic Field. : Consider a magnet of length 2l of POLE STRENGTH `q_(m)` kept in a uniform magnetic field `vec(B)`. Each pole experinences a force of magnitude `q_(m)` B but acts in opposite direction. Therefore, the net force exerted on the magnet is zero, so that there is no translatory motion. These two forces constitute a couple (about midpoint of bar magnet ) which will ROTATE and try to align in the direction of the magnetic field `vec(B)` . The force experienced by NORTH pole, `vec(F_(N)) = q_(m) vec(B)` The force `e^(x)` perienced by south pole, `vec(F_(S)) = - q_(m) vec(B)` Adding equations (1) and (2), we get the net force acting on the dipole as `vec(F ) = vec(F_(N)) = vec(0)` This implies , that the net force acting on the dipole is zero, but forms a couple which tends to rotate the ber magnet clockwise (here) in order to align it along `vec(B)`. The moment of force or torque experiencedby north and south pole about point O is `vec(tau) = vec(ON) xx vec(F_(N)) + vec(OS) xx vec(F_(s))` `vec(tau) = vec(ON) xx q_(m) vec(B) + vec(OS) xx (-q_(m) vec(B))`, By using right hand cork screw rule, we coclude that the total torque is pointing into the paper. Since the magnitudes `|vec(ON)|= |vec(OS)| = l and |q_(m) vec(B)| = |-q_(m) vec(B)|` The magnitude of total torque about point O `tau = l xx q_(m) B sin theta + l xx q_(m) B sin theta` `tau = 2l xx q_(m) B sin theta` `tau = p_(m) B sin theta "" (thetafore q_(m) xx 2l = p_(m))` In vector notation, `tau = p_(m) xx vec(B)` 
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