A charge of 4.0 C is moving with a velocity of 4.0 10^6 m/s along… — Solution | JEE Physics

MCQEasyJEE 2024Force on Moving Charge

A charge of $4.0 \, \mu \text{C}$ is moving with a velocity of $4.0 \times 10^6 \, \text{m/s}$ along the positive $y$ -axis under a magnetic field $B$ of strength $2\hat{k} \, \text{T}$ . The force acting on the charge is $x\hat{i} \, \text{N}$ . The value of $x$ is:

A
$32 \, \text{N}$
B
$24 \, \text{N}$
C
$20 \, \text{N}$
D
$16 \, \text{N}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $q = 4.0 \times 10^{-6} \, \text{C}$ , $\vec{v} = 4.0 \times 10^6 \, \hat{j} \, \text{m/s}$ , and $\vec{B} = 2\hat{k} \, \text{T}$ .

Find: The value of $x$ in $\vec{F} = x\hat{i} \, \text{N}$ .

Use the magnetic force relation:

\vec{F} = q(\vec{v} \times \vec{B})

The cross product is evaluated as:

\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4.0 \times 10^6 & 0 \\ 0 & 0 & 2 \end{vmatrix}

So,

\vec{v} \times \vec{B} = \hat{i}(4.0 \times 10^6 \times 2) - \hat{j}(0 \times 0) + \hat{k}(0 \times 0) = 8.0 \times 10^6 \hat{i}

Substituting into the force equation:

\vec{F} = (4.0 \times 10^{-6})(8.0 \times 10^6 \hat{i}) = 32\hat{i} \, \text{N}

Therefore, the value of $x$ is $32$ , and the correct option is A.

Common mistakes

Using $\vec{F} = q\vec{v}B$ without checking direction is incorrect because magnetic force is a vector cross product. Always evaluate $\vec{v} \times \vec{B}$ first to get both magnitude and direction.
Taking $\hat{j} \times \hat{k}$ with the wrong sign is incorrect. The right-hand rule gives $\hat{j} \times \hat{k} = \hat{i}$ , so the force is along positive $x$ -direction.
Forgetting to convert microcoulomb to coulomb leads to a large numerical error. Use $4.0 \, \mu \text{C} = 4.0 \times 10^{-6} \, \text{C}$ before substitution.

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