Verified Solution • 2025–26 Session moderate

QPages 230-234: Q34: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q34 Solution: A charged particle is accelerated through a potential…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the potential difference from kilovolts to volts, using $12$ instead of $12000$ in the calculation for $q/m$. This error leads to a radius that is 1000 times larger than the correct value, so always ensure all SI units are consistent before substituting into formulas.

A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of

1.0 \times 10^6 \text{ m s}^{-1}

. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.

Step-by-Step Verified Solution

**Answer:**

r = 0.5 \text{ m}

. **Step-by-Step Explanation:**

V = 12 \text{ kV}

,

v = 1.0 \times 10^6 \text{ m/s}

,

B = 0.2 \text{ T}

. From

\frac{1}{2} m v^2 = q V

,

\frac{q}{m} = \frac{v^2}{2V} = \frac{10^{12}}{24000} = 4.17 \times 10^7 \text{ C/kg}

.

r = \frac{v}{(q/m) B} = \frac{10^6}{4.17 \times 10^7 \times 0.2} = 0.12 \text{ m}

.

All Questions in this Chapter

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