AP EAMCET · PHYSICS · Magnetic Effects of Current
A small block of mass \(20 \mathrm{~g}\) and charge \(4 \mathrm{mC}\) is released on a long smooth inclined plane of inclination angle of \(45^{\circ}\), A uniform horizontal magnetic field of \(1 \mathrm{~T}\) is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is
- A \(2 \mathrm{~s}\)
- B \(3 \mathrm{~s}\)
- C \(5 \mathrm{~s}\)
- D \(6 \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Given inclination of plane, Mass of block, \(m=20 \mathrm{~g}=0.02 \mathrm{~kg}\) charge on the block, \(q=4 \mathrm{mC}=4 \times 10^{-3} \mathrm{C}\) Magnetic field, \(B=1 \mathrm{~T}\) Magnetic force on the charge particles, \(F=B q v\) Particle will leave the inclined plane,…
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