KCET · Physics · Magnetic Effects of Current
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
- A \(5.98 \mathrm{~A}\)
- B \(14.76 \mathrm{~A}\)
- C \(11.32 \mathrm{~A}\)
- D \(7.14 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(11.32 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
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,
\(F \cos 30^{\circ}=m g \sin 30^{\circ}\)
\(\Rightarrow B I l \cos 30^{\circ}=m g \sin 30^{\circ}\)
\(\Rightarrow B I l \times \frac{\sqrt{3}}{2}=m g \frac{1}{2} \Rightarrow \sqrt{3} B I=\left(\frac{m}{l}\right) g\)
\(I=\left(\frac{m}{l}\right) g \cdot \frac{1}{\sqrt{3} B}=0.5 \times 10 \times \frac{1}{\sqrt{3} \times 0.25}=11.32 \mathrm{~A}\)
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,
\(F \cos 30^{\circ}=m g \sin 30^{\circ}\)
\(\Rightarrow B I l \cos 30^{\circ}=m g \sin 30^{\circ}\)
\(\Rightarrow B I l \times \frac{\sqrt{3}}{2}=m g \frac{1}{2} \Rightarrow \sqrt{3} B I=\left(\frac{m}{l}\right) g\)
\(I=\left(\frac{m}{l}\right) g \cdot \frac{1}{\sqrt{3} B}=0.5 \times 10 \times \frac{1}{\sqrt{3} \times 0.25}=11.32 \mathrm{~A}\)
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