1μC charge moving with velocity… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Electromagnetism

2026integermedium

$1\,\mu$ C charge moving with velocity $\vec{v} = \left(\hat{i} - 2\hat{j} + 3\hat{k}\right)$ m/s in the region of magnetic field $\vec{B} = \left(2\hat{i} + 3\hat{j} - 5\hat{k}\right)$ T. The magnitude of force acting on it is $\sqrt{\alpha} \times 10^{-6}$ N. The value of $\alpha$ is _______.

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

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Solution

The magnetic force acting on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$ .

Given:

$q = 1 \, \mu\text{C} = 10^{-6} \text{ C}$

$\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}$

$\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}$

Calculating the cross product $\vec{v} \times \vec{B}$ :

$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$

$\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$

$\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}$

The magnitude of $\vec{v} \times \vec{B}$ is:

$|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$

The magnitude of the force is:

$|\vec{F}| = q |\vec{v} \times \vec{B}| = 10^{-6} \times \sqrt{171} = \sqrt{171} \times 10^{-6} \text{ N}$

Comparing this with the given expression $\sqrt{\alpha} \times 10^{-6} \text{ N}$ , we get:

$\alpha = 171$

Answer: $171$

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JEE Main 2026 — Physics Electromagnetism

2026integermedium

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The magnetic force acting on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$ .

Given:

$q = 1 \, \mu\text{C} = 10^{-6} \text{ C}$

$\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}$

$\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}$

Calculating the cross product $\vec{v} \times \vec{B}$ :

$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$

$\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$

$\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}$

The magnitude of $\vec{v} \times \vec{B}$ is:

$|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$

The magnitude of the force is:

$|\vec{F}| = q |\vec{v} \times \vec{B}| = 10^{-6} \times \sqrt{171} = \sqrt{171} \times 10^{-6} \text{ N}$

Comparing this with the given expression $\sqrt{\alpha} \times 10^{-6} \text{ N}$ , we get:

$\alpha = 171$

Answer: $171$

Your note

JEE Main Electromagnetism in other years

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Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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