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A 100 turn rectangular coil $\mathrm{ABCD}$ (in X-Yplane) is hung from one arm of a balance figure. A mass $500 \mathrm{~g}$ 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 $0.2 \mathrm{~T}$ acting inward (in $\mathrm{x}-\mathrm{z}$ plane) is switched on such that only arm CD of length $1 \mathrm{~cm}$ lies in the field. How much additional mass $m$ must be added to regain the balance?
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Verified Answer
The magnetic field is perpendicular to arms $\mathrm{BC}$ and $\mathrm{AD}$, so torque acts on $\mathrm{CD}$ and $\mathrm{AB}$ arms which rotate coil.
For equilibrium the net torque should also be equal to zero.
When the field is off $\Sigma t=0$ considering the separation of each hung from mid-point be $l$.
$$
\begin{aligned}
\mathrm{Mg} l &=\mathrm{W}_{\text {coil }} l \\
500 \mathrm{~g} l &=\mathrm{W}_{\text {coil }} l \\
\mathrm{~W}_{\text {coil }} &=500 \times 9.8 \mathrm{~N}
\end{aligned}
$$
Taking moment of force about mid-point, we get the weight of coil
When the magnetic field is switched on and the weight $(\mathrm{mg})$ is added on other side of beam balance to balance the coil,
$$
\begin{aligned}
&\begin{aligned}
\mathrm{Mg} l+\mathrm{mg} l &=\mathrm{W}_{\text {coil }} l+\mathrm{IBL} \sin 90^{\circ} \mathrm{I} \\
\mathrm{mg} l &=\mathrm{BIL} l
\end{aligned} \\
&\mathrm{~m}=\frac{\mathrm{BIL}}{\mathrm{g}}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3} \mathrm{~kg}=1 \mathrm{~g}
\end{aligned}
$$
Hence, I $g$ of additional mass must be added to regain the balance.
For equilibrium the net torque should also be equal to zero.
When the field is off $\Sigma t=0$ considering the separation of each hung from mid-point be $l$.
$$
\begin{aligned}
\mathrm{Mg} l &=\mathrm{W}_{\text {coil }} l \\
500 \mathrm{~g} l &=\mathrm{W}_{\text {coil }} l \\
\mathrm{~W}_{\text {coil }} &=500 \times 9.8 \mathrm{~N}
\end{aligned}
$$
Taking moment of force about mid-point, we get the weight of coil
When the magnetic field is switched on and the weight $(\mathrm{mg})$ is added on other side of beam balance to balance the coil,
$$
\begin{aligned}
&\begin{aligned}
\mathrm{Mg} l+\mathrm{mg} l &=\mathrm{W}_{\text {coil }} l+\mathrm{IBL} \sin 90^{\circ} \mathrm{I} \\
\mathrm{mg} l &=\mathrm{BIL} l
\end{aligned} \\
&\mathrm{~m}=\frac{\mathrm{BIL}}{\mathrm{g}}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3} \mathrm{~kg}=1 \mathrm{~g}
\end{aligned}
$$
Hence, I $g$ of additional mass must be added to regain the balance.
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