Question

$A$ rectangular coil $ABCD$ is hung from one side of a balance as shown in figure. $A$ $500\, g$ mass is added to the other arm to balance the weight of the coil. A current of $9.8 \,A$ is passed through the coil and a constant magnetic field of $0.4 \,T$ acting inward (in $xz$ plane) is switched on such that only arm $CD$ of length $1.5\, cm$ lies in the field. The additional mass $m$ must be added to regain the balance is

• A. $4 \, g$
• B. $5 \, g$
• C. $6 \, g$
• D. $7 \, g$

Answer 1

Answer: (C)

In absence of magnetic field the weight added in one pan balances the rectangular coil in the other pan of balance,
$\therefore Mgl = W_{coil} l$ or $ W_{coil} = Mg=0.5\times9.8 \,N$
When current $I$ is passed through the coil and the magnetic field is switched on
Let $m$ mass be added in the first pan to regain the balance Then $Mdl +mgl =W _{coil} l +IBL \, sin 90^{\circ}\,l$
$mgl = IBLI$
or $m=\frac{IBL}{g}=\frac{9.8\times0.4\times1.5\times10^{-2}}{9.8}$
$=0.6 \times10^{-2}\, kg$
$=6 \times10^{-3}\, kg =6\,g$