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Question
A 100 turn rectangular coil ABCD (in XY plane) is hung from one arm of a balance (Figure). A mass 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward (in xz plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass ‘m’ must be added to regain the balance?
Solution
Here we use the concept of magnetic force on straight current-carrying conductor placed in the region of external uniform magnetic field. The magnetic force exerted on CD due to external magnetic field must balance its weight.
And spring balance to be in equilibrium net torque should also be equal to zero.
At t = 0, the external magnetic field is off. Let us consider the separation of each hung from mid-point be `l`.
`Mgl = W_("coil")l`
`0.5 gl = W_("coil") l`
`W_("coil") = 0.5 xx 9.8 N`
By taking moment of force about mid-point, we get the weight of coil.
And Let `'m` be the mass which is added to regain the balance.
When the magnetic field is switched on.
`Mgl + mgl = W_("coil") l + (ILB sin 90^circ)l`
`Mgl = (ILB)l`
`m = (BIL)/g = (0.2 xx 4.9 xx 1 xx 10^-2)/9.8 = 10^-3 kg = 1g`
Therefore, 1 g of additional mass must be added to regain the balance.
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