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NCERT Exemplar · Q21

Q.A rectangular coil ABCD of 100100 turns hangs vertically from one arm of a beam balance, with its lower horizontal side CD (length 1 cm1\ \text{cm}) at the bottom. A 500 g500\ \text{g} mass placed on the other pan exactly balances the weight of the coil. A current of 4.9 A4.9\ \text{A} is then sent through the coil while a uniform magnetic field of 0.2 T0.2\ \text{T}, directed horizontally into the region and perpendicular to CD, is switched on so that only the arm CD lies in the field. Find the additional mass mm that must be added to the other pan to restore the balance.

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When current flows, the magnetic field pushes only on the horizontal arm CD, adding a vertical force F=NBIlF=NBIl to that side of the balance. To re-balance, a mass mm is added on the other pan so that mg=NBIlmg=NBIl; this gives m=0.1 kg=100 gm=0.1\ \text{kg}=100\ \text{g}.

Concept & formula

A straight current-carrying segment of length ll in a uniform field BB perpendicular to it feels a force F=BIlF=BIl per turn. For NN turns the force on the arm CD is

F=N B I l.F=N\,B\,I\,l.

The fields on the vertical arms and on the upper arm either cancel or lie outside the field region, so only CD contributes a net vertical force that disturbs the balance.

Step 1 — force on arm CD

F=NBIl=100×0.2 T×4.9 A×0.01 m=0.98 N.F=N B I l=100\times0.2\ \text{T}\times4.9\ \text{A}\times0.01\ \text{m}=0.98\ \text{N}.

Step 2 — additional mass to restore balance …

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