A particle of charge q and mass m is moving with a velocity -vi (v ≠ 0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field B = B₀k, the maximum value of v for which the particle will not hit the screen is: (a) (2qdB₀)/(m) (b) (qdB₀)/(3m) (c) (qdB₀)/(2m) (d) (qdB₀)/(m)

Answer: (d). v ⊥ B, so the particle moves in a circle of radius r = (mv)/(qB₀). Starting perpendicular to the screen, it gets closest to the screen after a quarter turn, a distance r further on. It does not hit the screen if r ≤ d: (mv)/(qB₀) ≤ d ⇒ v ≤ (qdB₀)/(m). So the answer is (d) (qdB₀)/(m).

A particle of charge q and mass m is moving with a velocity -vi (v ≠… | JEE Main 2020 (6 Sep, Shift 1)

Q81JEE Main 2020 · 6 SEP · SHIFT 1Difficult

Moving Charge in a Magnetic Field

A particle of charge q and mass m is moving with a velocity $-v\hat{i}$ (v $\neq$ 0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field $\vec{B} = B_0\hat{k}$ , the maximum value of v for which the particle will not hit the screen is:

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Answer: (d)

$\vec{v} \perp \vec{B}$ , so the particle moves in a circle of radius $r = \dfrac{mv}{qB_0}$ .
Starting perpendicular to the screen, it gets closest to the screen after a quarter turn, a distance $r$ further on.
It does not hit the screen if $r \le d$ : $\dfrac{mv}{qB_0} \le d \Rightarrow v \le \dfrac{qdB_0}{m}$ .
So the answer is (d) $\dfrac{qdB_0}{m}$ .

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Moving Charge in a Magnetic Field question from JEE Main 2020 (6 Sep, Shift 1)

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