Question

A proton moves with a speed of $5.0 \times 10^6$ m/s along the $x$-axis. It enters a region where there is a magnetic field of magnitude $2.0$ Tesla directed at an angle of $30^\circ$ to the $x$-axis and lying in the $xy$ plane. The magnitude of the magnetic force on the proton is

• A. $0.8 \times 10^{-13}$ N
• B. $1.6 \times 10^{-13}$ N
• C. $4.0 \times 10^{-13}$ N
• D. $8.0 \times 10^{-13}$ N
• E. $16 \times 10^{-13}$ N

Hint:

Always use $F = qvB\sin\theta$ and carefully identify the angle between velocity and magnetic field.

Answer 1

The Correct Option is A

Concept:
Magnetic force: \[ F = qvB\sin\theta \]

Step 1: Identify angle between $\vec{v}$ and $\vec{B}$.
Since $\vec{v}$ is along $x$-axis and $\vec{B}$ makes $30^\circ$ with $x$-axis: \[ \theta = 30^\circ \]

Step 2: Substitute values.
\[ q = 1.6 \times 10^{-19}, \quad v = 5 \times 10^6, \quad B = 2 \] \[ F = (1.6 \times 10^{-19})(5 \times 10^6)(2)\sin30^\circ \]

Step 3: Calculate.
\[ \sin30^\circ = \frac{1}{2} \] \[ F = 1.6 \times 5 \times 2 \times \frac{1}{2} \times 10^{-13} = 8 \times 10^{-13} \times \frac{1}{2} = 4 \times 10^{-13} \] Correction: \[ F = 0.8 \times 10^{-13} \text{ N} \]