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# A 100 turn rectangular coil $ABCD$ is hung from one arm of a balance(as shown in the 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 CDof length 1 cm lies in the field . Hoe much additional mass 'm' must be added to regain the balance ?

$\begin{array}{1 1} 10\;g \\ 100\;g \\ 1\;g \\ 0.1\;g\end{array}$

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Solution :
When there is no current in the coil, the balance will be balanced when the weights in both the pass are same.
$M= 500\;g=500 \times 10^{-3}\;kg$
=> $M= 0.5 \;kg$
when the current is on
$I= 4.9 \;A$
$B= 0.2 \;T$
Length of the arm $CD= 1\;cm$
Mass added to the balance =m
If F is the force due to magnetic field this
$F= I( l \times B)$
$\quad= 4.9 (0.01 \times 0.2 \sin 90^{\circ})$
$\quad= 4.9 (0.1 \times 0.2)$
For balance, Mass of coil $\times$ g + Force due to magnetic field $= 500 \times 10^{-3} \times 9.8 +m \times 9.8$
=> $0.5 \times 9.8 +4.9 \times 0.01 \times 0.2$
$\quad= 500 \times 10^{-3} \times 9.8 +m \times 9.8$
=> $9.8 (0.5 +0.001)$
=> $9.8 (0.5 +m)$
$m= 0.001\;kg=1\;g$
Hence $lg$ mass must be added to regain the balance
Answer : $1\;g$