A particle of charge q moves with a velocity V = ai in a magnetic field B = bj + ck, where ' a ', ' b ' and ' c ' are constants. The magnitude of force experienced by particle is (a) qa√(b² + c²) (b) qa(b + c) (c) qa√(b² - c²) (d) zero

Answer: (a). F = q( V× B) = q a i×(b j + c k) i× j = k, i× k = - j ⇒ F = qa(b k - c j) | F| = qa√(b² + c²) So the answer is (a).

A particle of charge q moves with a velocity V = ai in a magnetic… | MHT CET 2025 (22 Apr, Shift 2)

Q18MHT CET 2025 · 22 APR · SHIFT 2Medium

Moving Charge in a Magnetic Field

A particle of charge $q$ moves with a velocity $\vec{V} = a\hat{i}$ in a magnetic field $\vec{B} = b\hat{j} + c\hat{k}$ , where ' $a$ ', ' b ' and ' c ' are constants. The magnitude of force experienced by particle is

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Answer: (a)

$\vec F = q(\vec V\times\vec B) = q\,a\hat i\times(b\hat j + c\hat k)$
$\hat i\times\hat j = \hat k$ , $\hat i\times\hat k = -\hat j \Rightarrow \vec F = qa(b\hat k - c\hat j)$
$|\vec F| = qa\sqrt{b^2 + c^2}$
So the answer is (a).

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