Question

A positively charged particle \((q)\) travelling at \(30^\circ\) with respect to the direction of a magnetic field of strength \(2.4 \times 10^{-6}\,\text{T}\) experiences a force of \(4.8 \times 10^{-19}\,\text{N}\). The speed of the charged particle will be \([q = 1.6 \times 10^{-19}\,\text{C},\ \sin 30^\circ = \tfrac{1}{2}]\)

• A. \(5 \times 10^{6}\,\text{m s}^{-1}\)
• B. \(2.5 \times 10^{6}\,\text{m s}^{-1}\)
• C. \(2 \times 10^{6}\,\text{m s}^{-1}\)
• D. \(7.5 \times 10^{6}\,\text{m s}^{-1}\)

Hint:

Magnetic force depends on the component of velocity perpendicular to the magnetic field.

Answer 1

The Correct Option is B

Step 1: Write the formula for magnetic force.
The magnetic force on a charged particle is given by:
\[ F = qvB\sin\theta \]
Step 2: Substitute given values.
\[ 4.8 \times 10^{-19} = (1.6 \times 10^{-19}) \, v \, (2.4 \times 10^{-6}) \, \frac{1}{2} \]
Step 3: Simplify the expression.
\[ 4.8 \times 10^{-19} = 1.92 \times 10^{-25} \, v \]
Step 4: Solve for velocity.
\[ v = \frac{4.8 \times 10^{-19}}{1.92 \times 10^{-25}} = 2.5 \times 10^{6}\,\text{m s}^{-1} \]
Step 5: Conclusion.
The speed of the charged particle is \(2.5 \times 10^{6}\,\text{m s}^{-1}\).