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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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: $171$

JEE Main 2026 — Physics Electromagnetism

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Solution

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JEE Main 2026 — Physics Electromagnetism

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism