Magnetic Force on a Charged Particle Moving Northwest

Physics Problem: Magnetic Force on a Charged Particle Moving Northwest

Problem Statement

A charged particle with a charge of 10 nC is moving with a velocity of 15 m/s towards the northwest. Calculate the force on the particle due to the Earth's magnetic field in a place where the Earth's field is directed vertically upwards.

Given

Charge: \( q = 10 \, \text{nC} = 10 \times 10^{-9} \, \text{C} \)

Velocity: \( \vec{v} = 15 \, \text{m/s} \) (northwest, which is at a 45° angle between west and north)

Magnetic field: \( \vec{B} = 50 \, \mu\text{T} = 50 \times 10^{-6} \, \text{T} \) (upward, \(+\hat{k}\))

Solution

The magnetic force on a moving charge is given by:

\[ \vec{F} = q \vec{v} \times \vec{B} \]

Velocity vector (northwest):

\[ \vec{v} = 15 \cos(45^\circ) (-\hat{i}) + 15 \sin(45^\circ) (\hat{j}) = \frac{15}{\sqrt{2}} (-\hat{i} + \hat{j}) \]

Magnetic field vector: \( \vec{B} = 50 \times 10^{-6} \hat{k} \)

Cross product:

\[ \vec{v} \times \vec{B} = \frac{15}{\sqrt{2}} (-\hat{i} + \hat{j}) \times \left(50 \times 10^{-6} \hat{k}\right) \]

\[ = \frac{15}{\sqrt{2}} \cdot 50 \times 10^{-6} \left[(-\hat{i}) \times \hat{k} + \hat{j} \times \hat{k}\right] = \frac{15}{\sqrt{2}} \cdot 50 \times 10^{-6} (-\hat{j} - \hat{i}) \]

Magnitude:

\[ F = qvB\sin(90^\circ) = 10 \times 10^{-9} \cdot 15 \cdot 50 \times 10^{-6} = 7.5 \times 10^{-12} \, \text{N} \]

Direction: South-East (since the force vector is in the direction of \( -\hat{i} - \hat{j} \))

Discussion

This problem demonstrates how to apply vector decomposition and the right-hand rule to determine the magnetic force on a charged particle. Since the particle moves northwest and the magnetic field is upward, the force is perpendicular to both — resulting in a direction toward the southeast.

The magnitude of the force is small, but this example highlights the importance of vector analysis in understanding electromagnetic interactions.