Question

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

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

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 $W_{coil}=Mg=0.5\times 9.8N$
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 $Mdl+mgl=W_{coil}l+IBLsin90^{\circ }l$
$mgl=IBLI$
or $m=\frac{IBL}{g}=\frac{9.8\times 0.4\times 1.5\times 10^{-2}}{9.8}$
$=0.6\times 10^{-2}kg$
$=6\times 10^{-3}kg=6g$