A charge particle is moving in a uniform magnetic field (2i + 3j) T. If it has an acceleration of ( - 4j) m/s², then the value of α will be: (a) 3 (b) 6 (c) 12 (d) 2

Answer: (b). Magnetic force (and so acceleration) is always perpendicular to B: a· B = 0 (α i - 4 j)·(2 i + 3 j) = 2α - 12 = 0 α = 6 So the answer is (b) 6.

A charge particle is moving in a uniform magnetic field (2i + 3j) T… | JEE Main 2022 (26 Jul, Shift 1)

Q55JEE Main 2022 · 26 JUL · SHIFT 1Medium

Moving Charge in a Magnetic Field

A charge particle is moving in a uniform magnetic field $(2\hat{i} + 3\hat{j})\text{ T}$ . If it has an acceleration of $(\alpha\hat{i} - 4\hat{j})\text{ m/s}^2$ , then the value of $\alpha$ will be:

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Answer: (b)

Magnetic force (and so acceleration) is always perpendicular to $\vec B$ : $\vec a\cdot\vec B = 0$
$(\alpha\hat i - 4\hat j)\cdot(2\hat i + 3\hat j) = 2\alpha - 12 = 0$
$\alpha = 6$
So the answer is (b) 6.

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