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 force experienced by particle is

• A. \( qa\sqrt{b^2 + c^2} \)
• B. \( qa(b + c) \)
• C. \( qa\sqrt{b^2 - c^2} \)
• D. zero

Hint:

Force is maximum when velocity is perpendicular to the magnetic field.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A moving charge in a magnetic field experiences a force known as the Lorentz magnetic force.
This force is a vector quantity that is perpendicular to both the velocity vector and the magnetic field vector.
Step 2: Key Formula or Approach:
The magnetic force \( \vec{F} \) experienced by a particle of charge \( q \) moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by the cross product: \[ \vec{F} = q(\vec{v} \times \vec{B}) \] The magnitude of a vector \( \vec{A} = x\hat{i} + y\hat{j} + z\hat{k} \) is \( |\vec{A}| = \sqrt{x^2 + y^2 + z^2} \).
Step 3: Detailed Explanation:
Given the velocity vector: \( \vec{v} = a\hat{i} \)
Given the magnetic field vector: \( \vec{B} = b\hat{j} + c\hat{k} \)
First, we calculate the cross product \( \vec{v} \times \vec{B} \): \[ \vec{v} \times \vec{B} = (a\hat{i}) \times (b\hat{j} + c\hat{k}) \] Using the distributive property of the cross product: \[ \vec{v} \times \vec{B} = a\hat{i} \times b\hat{j} + a\hat{i} \times c\hat{k} \] Using the standard unit vector cross products (\( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{i} \times \hat{k} = -\hat{j} \)): \[ \vec{v} \times \vec{B} = ab(\hat{i} \times \hat{j}) + ac(\hat{i} \times \hat{k}) \] \[ \vec{v} \times \vec{B} = ab(\hat{k}) + ac(-\hat{j}) \] \[ \vec{v} \times \vec{B} = -ac\hat{j} + ab\hat{k} \] Now, multiply by the charge \( q \) to find the force vector \( \vec{F} \): \[ \vec{F} = q(-ac\hat{j} + ab\hat{k}) = -qac\hat{j} + qab\hat{k} \] We need to find the magnitude of this force vector, \( |\vec{F}| \): \[ |\vec{F}| = \sqrt{(-qac)^2 + (qab)^2} \] \[ |\vec{F}| = \sqrt{q^2 a^2 c^2 + q^2 a^2 b^2} \] Factor out the common term \( q^2 a^2 \) from under the square root: \[ |\vec{F}| = \sqrt{q^2 a^2 (c^2 + b^2)} \] Taking the square root of the squared terms: \[ |\vec{F}| = qa\sqrt{b^2 + c^2} \] This matches option (A).
Step 4: Final Answer:
The magnitude of force experienced by the particle is \( qa\sqrt{b^2 + c^2} \).