Question

A proton enters a uniform magnetic field $\vec{B} = 0.5 \hat{k} \ \text{T}$ with velocity $\vec{v} = 10^6 \hat{i} \ \text{m/s}$. The magnitude of the magnetic force on the proton is:

• A. $8.0 \times 10^{-14}$ N
• B. $1.6 \times 10^{-13}$ N
• C. $8.0 \times 10^{-13}$ N
• D. 0

Hint:

Magnetic force is maximum when velocity is perpendicular to the magnetic field. Use vector cross product rules.

Answer 1

The Correct Option is A

The magnetic force on a moving proton in a magnetic field is calculated using the formula for the magnetic force on a charged particle:

\[ F = qvB\sin\theta \]

Where:

• \( F \) represents the magnetic force.
• \( q \) is the proton's charge (\(1.6 \times 10^{-19} \ \text{C}\)).
• \( v \) is the proton's velocity (\(10^6 \ \text{m/s}\)).
• \( B \) is the magnetic field strength (\(0.5 \ \text{T}\)).
• \( \theta \) is the angle between the velocity and the magnetic field direction.

Given the velocity \(\vec{v} = 10^6 \hat{i} \ \text{m/s}\) (in the \( \hat{i} \) direction) and the magnetic field \(\vec{B} = 0.5 \hat{k} \ \text{T}\) (in the \( \hat{k} \) direction), the angle \(\theta\) between them is \(90^\circ\). Consequently, \(\sin\theta = \sin 90^\circ = 1\).

Substituting the values into the formula yields:

\[ F = (1.6 \times 10^{-19} \ \text{C})(10^6 \ \text{m/s})(0.5 \ \text{T})(1) \]

\[ F = 8.0 \times 10^{-14} \ \text{N} \]

Therefore, the magnitude of the magnetic force acting on the proton is \(8.0 \times 10^{-14} \ \text{N}\).