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A magnet is hung horizontally in the magnetic meridian by a wire without any twist. If the supporting wire is given a twist of 180° at the top, the magnet rotates by 30°. Now if another magnet is used, then a twist of 270° at the supporting end of wire also produces a rotation of the magnet by 30°. Compare the magnetic dipole moments of the two magnets. |
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Answer» Solution :If resultant TWIST in the wire `=delta`, `delta_(1) = 180^@ - 30^@ = 150^@ = 150 xx (pi)/( 180) "rad" and` `delta_(2) = 270^(@) - 30^(@) = 240^(@) xx (pi)/( 180) "rad"` If the twist - CONSTANT for the wire is k then Rotating torque `tau_(1) = k delta_(1) and tau_(2) = kdelta_(2)` `therefore (tau_(1) )/( tau_(2) ) = (delta_(1) )/( delta_(2))` Here `alpha` is the angle made by the magnetic dipole moment with the magnetic meridian `tau_(1) .= m_(1) B_(H) sin alpha` SINCE the second magnet is also rotated by the same angle `tau_(2). = m_(2) B_(H) sin alpha` `therefore (tau_(1) .) /( tau_(2) .) = (m_1)/( m_2)` At equilibrium, `tau_(1) = tau_(1). and tau_(2) = tau_(2).` `(tau_(1).) /( tau_(2) .) = (tau_(1) )/( tau_(2) )` `(m_1)/( m_2) = (delta_1)/(delta_2) = (150)/(240) = (5)/(8)` |
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