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

To determine the magnitude of the magnetic force on a proton moving in a magnetic field, we use the formula for the magnetic force exerted on a charged particle:

\[ F = qvB\sin\theta \]

where:

• \( F \) is the magnetic force,
• \( q \) is the charge of the proton (\(1.6 \times 10^{-19} \ \text{C}\)),
• \( v \) is the velocity of the proton (\(10^6 \ \text{m/s}\)),
• \( B \) is the magnetic field strength (\(0.5 \ \text{T}\)),
• \( \theta \) is the angle between the velocity and the magnetic field direction.

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

Substitute the values into the formula:

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

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

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

Answer 2

We use the magnetic force formula: \[ \vec{F} = q (\vec{v} \times \vec{B}) \] Given: \[ \vec{v} = 10^6 \hat{i} \text{ m/s}, \quad \vec{B} = 0.5 \hat{k} \text{ T} \] \[ \vec{v} \times \vec{B} = 10^6 \hat{i} \times 0.5 \hat{k} = -5 \times 10^5 \hat{j} \] \[ |\vec{F}| = q v B \sin \theta = 1.6 \times 10^{-19} \times 10^6 \times 0.5 = 8.0 \times 10^{-14} \text{ N} \] Answer: $\boxed{8.0 \times 10^{-14} \text{ N}}$