Rankers Physics

Force Acting on Moving Charges: Practice Problem & Solution

A charge having $q/m$ equal to $10^8\text{ C/kg}$ and with velocity $3 \times 10^5\text{ m/s}$ enters into a uniform magnetic field $B = 0.3\text{ tesla}$ at an angle $30^\circ$ with direction of field. Then radius of curvature will be: (2000)
$0.01\text{ cm}$
$0.5\text{ cm}$
$1\text{ cm}$
$2\text{ cm}$

Solution Explained:

To solve this problem, we apply the core principles of Force Acting on Moving Charges. Understanding the underlying formula is key to arriving at the correct answer below:

Radius is $r = \frac{vsin\theta}{(q/m)B}$. Substituting values yields $r = \frac{3 \times 10^5 sin(30^\circ)}{10^8 \times 0.3} = 0.005\text{ m} = 0.5\text{ cm}$.

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