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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2026 · 2 April · Morning Shift · Physics · Magnetic Effects of Current

1\,μ C charge moving with velocity v = ( i - 2 j + 3 k ) m/s in the...

1 μ1\,\muC charge moving with velocity v⃗=(i^−2j^+3k^)\vec{v} = \left(\hat{i} - 2\hat{j} + 3\hat{k}\right) m/s in the region of magnetic field B⃗=(2i^+3j^−5k^)\vec{B} = \left(2\hat{i} + 3\hat{j} - 5\hat{k}\right) T. The magnitude of force acting on it is α×10−6\sqrt{\alpha} \times 10^{-6} N. The value of α\alpha is _______.

Correct answer

0

Step-by-step solution

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

Given:
q=1 μC=10−6 Cq = 1 \, \mu\text{C} = 10^{-6} \text{ C}
v⃗=i^−2j^+3k^\vec{v} = \hat{i} - 2\hat{j} + 3\hat{k}
B⃗=2i^+3j^−5k^\vec{B} = 2\hat{i} + 3\hat{j} - 5\hat{k}

Calculating the cross product v⃗×B⃗\vec{v} \times \vec{B}:
v⃗×B⃗=∣i^j^k^1−2323−5∣\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}

v⃗×B⃗=i^(10−9)−j^(−5−6)+k^(3+4)\vec{v} \times \vec{B} = \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)
v⃗×B⃗=i^+11j^+7k^\vec{v} \times \vec{B} = \hat{i} + 11\hat{j} + 7\hat{k}

The magnitude of v⃗×B⃗\vec{v} \times \vec{B} is:
∣v⃗×B⃗∣=12+112+72=1+121+49=171|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}

The magnitude of the force is:
∣F⃗∣=q∣v⃗×B⃗∣=10−6×171=171×10−6 N|\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 α×10−6 N\sqrt{\alpha} \times 10^{-6} \text{ N}, we get:
α=171\alpha = 171

Answer: 171171
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