Verified Solution • 2025–26 Session moderate

QPages 230-234: Q52: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q52 Solution: An electron is emitted with negligible speed from the negative…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the maximum vertical displacement of a charged particle entering a magnetic field perpendicularly with the radius $r$, when it is actually the diameter $2r$. The condition for the electron to fail to strike the upper plate is $d > 2r$, not $d > r$.

An electron is emitted with negligible speed from the negative plate of a parallel-plate capacitor charged to a potential difference

V

. The separation between the plates is

d

and a magnetic field

B

exists in the space, as shown in the figure. Show that the electron will fail to strike the upper plates if

d > \sqrt{2mV/eB^2}

.

Step-by-Step Verified Solution

**Answer:**

d > \sqrt{\frac{2 m V}{e B^2}}

. **Step-by-Step Explanation:** The electron follows a circular path of radius

r = \frac{m v}{e B}

. The condition for it to not strike the upper plate is

d > 2r

. Substituting

v = \sqrt{2eV/m}

gives the result.

All Questions in this Chapter

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