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QPages 230-234: Q38: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q38 Solution: A particle of mass and positive charge , moving with a uniform…

⚠️ Common Exam Traps & Conceptual Pitfalls

When applying $r=\frac{mv}{qB}$ remember that $r$ is a magnitude; use $|q|$ in the denominator so the radius stays positive. The sign of $q$ only determines the sense of curvature, not the size of the arc, angle, or travel time.

A particle of mass mm and positive charge qq, moving with a uniform velocity vv, enters a magnetic field BB, as shown in the figure. a) Find the radius of the circular arc it describes in the magnetic field. b) Find the angle subtended by the arc at the centre. c) How long does the particle stay inside the magnetic field? d) Solve the three parts of the above problem if the charge qq on the particle is negative.

Step-by-Step Verified Solution

**Answer:** (a) r=mvqBr = \frac{m v}{q B} (b) θ=π−2θ\theta = \pi - 2\theta (c) t=m(π−2θ)qBt = \frac{m(\pi - 2\theta)}{q B} (d) Same for negative charge. **Step-by-Step Explanation:** (a) The radius of the circular arc is r=mvqBr = \frac{m v}{q B}. (b) The angle subtended at the centre is π−2θ\pi - 2\theta. (c) Time inside the field: t=m(π−2θ)qBt = \frac{m(\pi - 2\theta)}{q B}. (d) For a negative charge, the direction of curvature reverses, but the radius, angle, and time have the same magnitudes.

All Questions in this Chapter

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