Verified Solution • 2025–26 Session moderate

QPages 230-234: Q51: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q51 Solution: A particle of mass and charge is released from the origin in a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the magnetic field contributes to the particle’s kinetic energy and thus to its speed. In fact, the Lorentz force $q,vec v imesvec B$ is always perpendicular to $vec v$, so it changes only the direction of motion, not the magnitude. Therefore the speed is determined solely by the work done by the electric field, giving $\tfrac12 m v^2 = qE_0 z$ (with the sign of $q$ determining whether $z$ increases or decreases).

A particle of mass mm and charge qq is released from the origin in a region in which the electric field and magnetic field are given by B⃗=−B0j^\vec{B} = -B_0 \hat{j} and E⃗=E0k^\vec{E} = E_0 \hat{k}. Find the speed of the particle as a function of its zz-coordinate.

Step-by-Step Verified Solution

**Answer:** v=2qE0zmv = \sqrt{\frac{2 q E_0 z}{m}}. **Step-by-Step Explanation:** The electric field accelerates the particle along z, while the magnetic field causes circular motion in the xy-plane. The speed as a function of z is found from energy conservation.

All Questions in this Chapter

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