JEE Main 2025 — Circular Motion Question with Solution

To solve this problem, note that when a charged particle with mass $$m$$ and charge $$q$$ is fired perpendicular to a magnetic field of strength $$B$$, it undergoes uniform circular motion. The period (time to complete one full circle) is given by:

$$T=\frac{2\pi m}{qB}$$

Here's how to calculate it step by step:

Convert the given quantities to SI units:

Charge: $$q=1.6\mu \text{C}=1.6\times 10^{-6}\text{C}$$

Mass: $$m=16\mu \text{g}=16\times 10^{-9}\text{kg}$$

Magnetic field: $$B=6.28\text{T}$$

Write down the period formula:

$$T=\frac{2\pi m}{qB}$$

Substitute the values:

$$T=\frac{2\pi \times (16\times 10^{-9}\text{kg})}{(1.6\times 10^{-6}\text{C})(6.28\text{T})}$$

Notice that $$2\pi \approx 6.28$$, which cancels with the given magnetic field value, simplifying the expression:

$$T=\frac{6.28\times 16\times 10^{-9}}{1.6\times 10^{-6}\times 6.28}=\frac{16\times 10^{-9}}{1.6\times 10^{-6}}$$

Simplify the fraction:

$$T=\frac{16}{1.6}\times \frac{10^{-9}}{10^{-6}}=10\times 10^{-3}=0.01\text{s}$$

Thus, the time required for the particle to return to its original location is:

$$\boxed{0.01\text{s}}$$