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

Given: q = 4.0 10^-6 C, v = 4.0 10^6 m/s j, and B = 2 k T. Find: The value of x in F = x i N. Use the magnetic force relation F = q( v B) Now evaluate the cross product using the determinant form: v B = vmatrix i & j & k 0 & 4.0 10^6 & 0 0 & 0 & 2 vmatrix So, v B = i(4.0 10^6 2) = 8.0 10^6 i Substituting into the force equation, F = (4.0 10^-6)(8.0 10^6 i) F = 32 i N Therefore, the value of x is 32 and the correct option is A.

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 $\vec{B}$ 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: $q = 4.0 \times 10^{-6} \, \text{C}$ , $\vec{v} = 4.0 \times 10^6 \, \text{m/s} \, \hat{j}$ , 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})

Now evaluate the cross product using the determinant form:

\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) = 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})

\vec{F} = 32\hat{i} \, \text{N}

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

Common mistakes

Using $F = qvB$ without checking the direction. The magnitude formula gives the size of the force, but the question also specifies the force along $\hat{i}$ . Always use $\vec{F} = q(\vec{v} \times \vec{B})$ first, then identify both magnitude and direction.
Making an error in the right-hand rule for $\hat{j} \times \hat{k}$ . Here, $\hat{j} \times \hat{k} = \hat{i}$ , not $-\hat{i}$ . Use the cyclic order $\hat{i} \to \hat{j} \to \hat{k}$ to get the correct direction.
Forgetting to convert microcoulomb into coulomb. The charge is $4.0 \, \mu \text{C} = 4.0 \times 10^{-6} \, \text{C}$ . If this conversion is missed, the numerical answer becomes incorrect by a factor of $10^6$ .

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