Question

A particle of charge $Q$ moves with a velocity $\vec{v}=a\hat{i}$ in a magnetic field $\vec{B}=b\hat{j}+c\hat{k}$, where $a, b$ and $c$ are constants. The magnitude of the force experienced by the particle is

• A. $Qa(b+c)$
• B. Zero
• C. $Qa\sqrt{b^2 + c^2}$
• D. $Qa\sqrt{b^2 - c^2}$
• E. $Qa(b-c)$

Hint:

The magnetic force is always perpendicular to both the velocity and the magnetic field. Since the velocity is along the x-axis and the field is in the yz-plane, the entire field contributes to the force.

Answer 1

The Correct Option is C

Concept:
The magnetic force $\vec{F}$ acting on a charge $Q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force law: \[ \vec{F} = Q(\vec{v} \times \vec{B}) \]

Step 1: Calculate the cross product $\vec{v} \times \vec{B}$.
Given $\vec{v} = a\hat{i}$ and $\vec{B} = b\hat{j} + c\hat{k}$: \[ \vec{v} \times \vec{B} = (a\hat{i}) \times (b\hat{j} + c\hat{k}) \] Using the properties of unit vector cross products ($\hat{i} \times \hat{j} = \hat{k}$ and $\hat{i} \times \hat{k} = -\hat{j}$): \[ \vec{v} \times \vec{B} = a \cdot b (\hat{i} \times \hat{j}) + a \cdot c (\hat{i} \times \hat{k}) \] \[ \vec{v} \times \vec{B} = ab\hat{k} - ac\hat{j} \]

Step 2: Calculate the magnitude of the force.
The force vector is $\vec{F} = Q(ab\hat{k} - ac\hat{j}) = -Qac\hat{j} + Qab\hat{k}$. The magnitude $|\vec{F}|$ is: \[ |\vec{F}| = \sqrt{(-Qac)^2 + (Qab)^2} \] \[ |\vec{F}| = \sqrt{Q^2a^2c^2 + Q^2a^2b^2} \] \[ |\vec{F}| = Qa\sqrt{c^2 + b^2} \]