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:

Magnetic force follows the vector rule $\vec{F}=q(\vec{v}\times\vec{B})$. Try placing the velocity and field along two perpendicular axes first; this makes the cross product easy to evaluate and also tells you the direction of the force, not just its size.

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. \[ F = 9.6 \times 10^{-13}\,N \]

Answer 2

Concept:

• The magnetic force on a moving charge is a vector quantity given by the Lorentz force law $\vec{F} = q(\vec{v} \times \vec{B})$; the familiar scalar formula $qvB\sin\theta$ is just the magnitude of this cross product.
• Placing the velocity and magnetic field along two perpendicular coordinate axes makes the perpendicular condition automatic and lets the force be computed directly using vector identities instead of remembering to substitute $\sin 90^\circ$.

Step 1: Set up the velocity and field as perpendicular vectors along the axes.
Since the proton enters perpendicular to the field, take:
$\vec{v} = v\,\hat{x} = (3\times10^6)\,\hat{x}\ \text{m/s}$
$\vec{B} = B\,\hat{y} = (2)\,\hat{y}\ \text{T}$

Step 2: Compute the cross product $\vec{v} \times \vec{B}$.
$\vec{v} \times \vec{B} = (v\,\hat{x}) \times (B\,\hat{y}) = vB\,(\hat{x} \times \hat{y}) = vB\,\hat{z}$
using the standard identity $\hat{x} \times \hat{y} = \hat{z}$.

Step 3: Substitute the numeric values to get the magnitude of $\vec{v} \times \vec{B}$.
$|\vec{v} \times \vec{B}| = vB = (3\times10^6)(2) = 6\times10^6$

Step 4: Multiply by the charge to get the force.
$F = q\,|\vec{v} \times \vec{B}| = (1.6\times10^{-19})(6\times10^6)$
$F = 9.6\times10^{-13}\ N$

Final Answer: $F = 9.6\times10^{-13}\ N$, directed along $\hat{z}$, perpendicular to both $\vec{v}$ and $\vec{B}$