Physics Problem: Magnetic Force on a Copper Rod Moving Southeast
Problem Statement
In a laboratory setup, a copper rod is charged with 50 nC and is thrown with a horizontal velocity of 20 m/s towards the southeast. Calculate the force on the rod due to the Earth's magnetic field in a location where the Earth's magnetic field is directed vertically downwards.
Given
- Charge: \( q = 50 \, \text{nC} = 50 \times 10^{-9} \, \text{C} \)
- Velocity: \( \vec{v} = 20 \, \text{m/s} \) (southeast, at 45° between east and south)
- Magnetic field: \( \vec{B} = 50 \, \mu\text{T} = 50 \times 10^{-6} \, \text{T} \) (downward, \( -\hat{k} \))
Solution
The magnetic force on a moving charge is given by:
\[ \vec{F} = q \vec{v} \times \vec{B} \]
Velocity vector (southeast):
\[ \vec{v} = 20 \cos(45^\circ) \hat{i} + 20 \sin(45^\circ) (-\hat{j}) = \frac{20}{\sqrt{2}} (\hat{i} - \hat{j}) \]
Magnetic field vector: \( \vec{B} = -50 \times 10^{-6} \hat{k} \)
Cross product:
\[ \vec{v} \times \vec{B} = \frac{20}{\sqrt{2}} (\hat{i} - \hat{j}) \times (-50 \times 10^{-6} \hat{k}) \]
\[ = \frac{20}{\sqrt{2}} \cdot 50 \times 10^{-6} \left[ \hat{i} \times (-\hat{k}) - \hat{j} \times (-\hat{k}) \right] = \frac{20}{\sqrt{2}} \cdot 50 \times 10^{-6} (\hat{j} + \hat{i}) \]
Magnitude:
\[ F = qvB\sin(90^\circ) = 50 \times 10^{-9} \cdot 20 \cdot 50 \times 10^{-6} = 5.0 \times 10^{-11} \, \text{N} \]
Direction: Northeast (since the force vector is in the direction of \( \hat{i} + \hat{j} \))
Diagram
Discussion
This problem demonstrates how to apply the Lorentz force law to a charged object moving diagonally through a vertical magnetic field. The velocity vector is decomposed into components, and the cross product with the magnetic field vector gives the direction of the resulting force.
In this case, the rod moves southeast and the magnetic field is downward, so the force is directed northeast. The magnitude of the force is small, but it illustrates the interaction between motion, charge, and magnetic fields in three dimensions.
0 Comments