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

Concept: The magnetic force acting on a moving charged particle in a magnetic field is given by \[ F = qvB\sin\theta \] where \(q\) = charge of the particle, \(v\) = velocity of the particle, \(B\) = magnetic field strength, \(\theta\) = angle between velocity and magnetic field.

Step 1: Identify the given values. \[ q = 1.6 \times 10^{-19}\,C \] \[ v = 3 \times 10^6\,\text{m/s} \] \[ B = 2\,T \] Since the particle enters perpendicular to the field, \[ \theta = 90^\circ \] \[ \sin 90^\circ = 1 \]

Step 2: Substitute into the formula. \[ F = qvB\sin\theta \] \[ F = (1.6 \times 10^{-19})(3 \times 10^6)(2) \]

Step 3: Calculate the force. \[ F = 9.6 \times 10^{-13}\,N \]