Question

A proton enters into a magnetic field of intensity \( 5 \times 10^{-2} \) Tesla with velocity \( 10^5 \, \text{m/s} \) at an angle of 30° with the field. Calculate the magnitude of force acting on the proton due to this field.

Hint:

The magnetic force on a moving charged particle is given by \( F = q v B \sin \theta \), where \( \theta \) is the angle between the velocity of the particle and the magnetic field.

Answer 1

Step 1: Formula for magnetic force.
The magnetic force \( F \) on a charged particle moving with velocity \( v \) in a magnetic field \( B \) at an angle \( \theta \) to the magnetic field is given by the equation: \[ F = q v B \sin \theta \] where:
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field.
Step 2: Substituting the known values.
For a proton, the charge \( q = 1.6 \times 10^{-19} \, \text{C} \), the velocity \( v = 10^5 \, \text{m/s} \), the magnetic field strength \( B = 5 \times 10^{-2} \, \text{T} \), and the angle \( \theta = 30^\circ \). Substituting these values into the formula: \[ F = (1.6 \times 10^{-19}) \times (10^5) \times (5 \times 10^{-2}) \times \sin 30^\circ \] \[ F = (1.6 \times 10^{-19}) \times (10^5) \times (5 \times 10^{-2}) \times \frac{1}{2} \] \[ F = 4 \times 10^{-17} \, \text{N} \] Step 3: Conclusion.
Thus, the magnitude of the force acting on the proton due to the magnetic field is \( 4 \times 10^{-17} \, \text{N} \).