CUET · PHYSICS · PYQ PAPER 2023
A proton of charge \(1.6 \times 10^{-19} C\) is moving with a velocity of \(5 \times 10^3 m / s\) in a magnetic field of 4 T. The work done by the magnetic field on the proton is :
- A zero
- B \(8 \times 10^{-16} J\)
- C \(8 \times 10^{16} J\)
- D \(3.2 \times 10^{-16} J\)
Answer & Solution
Correct Answer
(A) zero
Step-by-step Solution
Detailed explanation
The magnetic force \(\vec{F}=q(\vec{v} \times \vec{B})\) is always perpendicular to the velocity of the proton. Since work done is defined by the dot product of force and displacement ( \(W=\vec{F} \cdot \vec{d}\) ), and the angle between them is always \(90^{\circ}\), the…
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