Question

A proton \((q = 1.6 \times 10^{-19}\,C)\) enters a magnetic field of \(2\,T\) with a velocity of \(3 \times 10^6\,\text{m/s}\) perpendicular to the field. Find the magnetic force acting on the proton.

• A. \(9.6 \times 10^{-14}\,N\)
• B. \(9.6 \times 10^{-13}\,N\)
• C. \(9.6 \times 10^{-12}\,N\)
• D. \(9.6 \times 10^{-11}\,N\)

Hint:

When a charged particle moves perpendicular to a magnetic field, \(\sin 90^\circ = 1\). So the force simplifies to \(F = qvB\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A charged particle (proton) is moving through a magnetic field at a right angle (perpendicular) to the field lines.
We must calculate the magnitude of the magnetic Lorentz force acting on it.
Step 2: Key Formula or Approach:
The magnetic force (\(F\)) experienced by a moving charged particle is given by the formula:
\[ F = qvB\sin\theta \] where \(q\) is charge, \(v\) is velocity, \(B\) is magnetic field strength, and \(\theta\) is the angle between velocity and the magnetic field.
Step 3: Detailed Solution:
Given values from the problem statement are:
Charge of proton, \(q = 1.6 \times 10^{-19}\,C\)
Velocity, \(v = 3 \times 10^6\,\text{m/s}\)
Magnetic field strength, \(B = 2\,T\)
Since the proton enters perpendicular to the magnetic field, the angle \(\theta = 90^\circ\).
We know that \(\sin(90^\circ) = 1\).
Substitute these values into the force formula:
\[ F = (1.6 \times 10^{-19}) \times (3 \times 10^6) \times (2) \times \sin(90^\circ) \] \[ F = (1.6 \times 3 \times 2) \times (10^{-19} \times 10^6) \times 1 \] \[ F = (9.6) \times 10^{-19 + 6} \] \[ F = 9.6 \times 10^{-13}\,N \] Step 4: Final Answer:
The magnetic force acting on the proton is \(9.6 \times 10^{-13}\,N\).