Question

The magnetic force acting on a charged particle carrying a charge of \( 3 \, \mu C \) in a magnetic field of 5 T acting in the \( y \)-direction, when the particle velocity is \( ( \hat{i} + \hat{j} ) \times 10^5 \, \text{ms}^{-1} \), is:

• A. 0.5 N in \( +x \) direction
• B. 0.2 N in \( +y \) direction
• C. 2 N in \( -x \) direction
• D. 1.5 N in \( +z \) direction
• E. 1 N in \( +z \) direction

Hint:

When calculating the magnetic force, remember that the force is always perpendicular to both the velocity of the particle and the direction of the magnetic field. Use the right-hand rule for cross products.

Answer 1

The Correct Option is D

Step 1: Recall the formula for magnetic force.}
The magnetic force on a charged particle moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by: \[ \vec{F} = q \, \vec{v} \times \vec{B} \] where: - \( \vec{F} \) is the magnetic force, - \( q \) is the charge on the particle, - \( \vec{v} \) is the velocity of the particle, - \( \vec{B} \) is the magnetic field.
Step 2: Write the given values.}
We are given: - \( q = 3 \, \mu C = 3 \times 10^{-6} \, \text{C} \), - \( \vec{v} = ( \hat{i} + \hat{j} ) \times 10^5 \, \text{ms}^{-1} \), - \( \vec{B} = 5 \, \text{T} \, \hat{j} \) (since the magnetic field is in the \( y \)-direction).
Step 3: Compute the cross product.}
Now, we calculate the cross product \( \vec{v} \times \vec{B} \): \[ \vec{v} = ( \hat{i} + \hat{j} ) \times 10^5 \] \[ \vec{B} = 5 \, \hat{j} \] The cross product is: \[ \vec{v} \times \vec{B} = \left( ( \hat{i} + \hat{j} ) \times 10^5 \right) \times 5 \hat{j} \] Using the properties of cross products and unit vectors: \[ \hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{j} = 0, \quad \hat{i} \times \hat{j} = \hat{k} \] We get: \[ \vec{v} \times \vec{B} = 5 \times 10^5 \left( \hat{i} \times \hat{j} \right) \] \[ \vec{v} \times \vec{B} = 5 \times 10^5 \hat{k} \]
Step 4: Calculate the magnetic force.}
Now, the magnetic force \( \vec{F} \) is: \[ \vec{F} = q \, \vec{v} \times \vec{B} = (3 \times 10^{-6}) \times 5 \times 10^5 \hat{k} \] \[ \vec{F} = 1.5 \, \text{N} \, \hat{k} \] Since \( \hat{k} \) represents the \( +z \)-direction, the force is \( 1.5 \, \text{N} \) in the \( +z \)-direction.