Question

One proton enters in a magnetic field of \( 2500 \, \text{N/Amp-m} \) intensity with velocity of \( 4 \times 10^5 \, \text{m/sec} \) in parallel of the field. The force exerted on the proton will be:

• A. 0 N
• B. \( 4.8 \times 10^{-10} \, \text{N} \)
• C. \( 0.48 \times 10^{-10} \, \text{N} \)
• D. \( 4.8 \times 10^{-10} \, \text{N} \)

Hint:

When a charged particle moves parallel or antiparallel to a magnetic field, the force on it is zero because the angle between the velocity and magnetic field is \( 0^\circ \) or \( 180^\circ \), and \( \sin 0^\circ = 0 \).

Answer 1

The Correct Option is A

The magnetic force on a proton is calculated using: \[ F = qvB \sin \theta \] Where: - \( q \) (proton charge) \( = 1.6 \times 10^{-19} \, \text{C} \) - \( v \) (velocity) \( = 4 \times 10^5 \, \text{m/s} \) - \( B \) (magnetic field intensity) \( = 2500 \, \text{N/Amp-m} \) - \( \theta = 0^\circ \) (proton moves parallel to the field) Since \( \sin 0^\circ = 0 \), the force is: \[ F = 1.6 \times 10^{-19} \times 4 \times 10^5 \times 2500 \times 0 = 0 \, \text{N} \] The force on the proton is therefore 0 N.