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A metallic rod of mass per unit length $0.5 \mathrm{~kg} \mathrm{~m}^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. A magnetic field of strength $0.25 \mathrm{~T}$ is acting on it in the vertical direction. When a current $I$ is flowing through it, the rod is not allowed to slide down. The quantity of current required to keep the rod stationary is
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Verified Answer
The correct answer is:
$11.32 \mathrm{~A}$
Given, magnetic field, $B=0.25 \mathrm{~T}$
Mass per unit length, $\frac{m}{l}=0.5 \mathrm{~kg} \mathrm{~m}^{-1}$
$$
\theta=30^{\circ}
$$
The given situation is shown alongside.
At balance condition,
$$
F \cos 30^{\circ}=m g \sin 30^{\circ}
$$
$\begin{aligned} & \Rightarrow \quad B I l \cos 30^{\circ}=m g \sin 30^{\circ} \\ & \Rightarrow \quad B I l \times \frac{\sqrt{3}}{2}=m g \frac{1}{2} \Rightarrow \sqrt{3} B I=\left(\frac{m}{l}\right) g\end{aligned}$
$$
I=\left(\frac{m}{l}\right) g \cdot \frac{1}{\sqrt{3} B}=0.5 \times 10 \times \frac{1}{\sqrt{3} \times 0.25}=11.32 \mathrm{~A}
$$
Mass per unit length, $\frac{m}{l}=0.5 \mathrm{~kg} \mathrm{~m}^{-1}$
$$
\theta=30^{\circ}
$$
The given situation is shown alongside.
At balance condition,
$$
F \cos 30^{\circ}=m g \sin 30^{\circ}
$$
$\begin{aligned} & \Rightarrow \quad B I l \cos 30^{\circ}=m g \sin 30^{\circ} \\ & \Rightarrow \quad B I l \times \frac{\sqrt{3}}{2}=m g \frac{1}{2} \Rightarrow \sqrt{3} B I=\left(\frac{m}{l}\right) g\end{aligned}$
$$
I=\left(\frac{m}{l}\right) g \cdot \frac{1}{\sqrt{3} B}=0.5 \times 10 \times \frac{1}{\sqrt{3} \times 0.25}=11.32 \mathrm{~A}
$$
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