The region between y = 0 and y = d contains a magnetic field B = Bz. A particle of mass m and charge q enters the region with a velocity v = vi. If d = mv/2qB, the acceleration of the charged particle at the point of its emergence at the other side is: (a) qvB/m(-√(3)/2i - 1/2j) (b) qvB/m(1/2i - √(3)/2j) (c) qvB/m((-j+i)/(√(2))) (d) qvB/m((j+i)/(√(2)))

Answer: (a). Radius of path: r = mv/qB = 2d. After turning through angle θ, the sideways shift is r(1 - cos θ) = d ⇒ cos θ = 1/2, θ = 60°. With F = qvi × Bk = -qvBj at entry, the velocity at emergence is v(cos 60 - sin 60) and the acceleration points to the centre, perpendicular to it. a = qvB/m(-sin 60 - cos 60) = qvB/m(-√(3)/2i - 1/2j) So the answer is (a).

The region between y = 0 and y = d contains a magnetic field B = Bz… | JEE Main 2019 (11 Jan, Shift 2)

Q93JEE Main 2019 · 11 JAN · SHIFT 2Difficult

Moving Charge in a Magnetic Field

The region between y = 0 and y = d contains a magnetic field $\vec{B} = B\hat{z}$ . A particle of mass m and charge q enters the region with a velocity $\vec{v} = v\hat{i}.$ If d $= \frac{mv}{2qB},$ the acceleration of the charged particle at the point of its emergence at the other side is:

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Answer: (a)

Radius of path: $r = \dfrac{mv}{qB} = 2d$ .
After turning through angle $\theta$ , the sideways shift is $r(1 - \cos\theta) = d \Rightarrow \cos\theta = \tfrac{1}{2}$ , $\theta = 60^\circ$ .
With $\vec{F} = qv\hat{i} \times B\hat{k} = -qvB\hat{j}$ at entry, the velocity at emergence is $v(\cos 60^\circ\hat{i} - \sin 60^\circ\hat{j})$ and the acceleration points to the centre, perpendicular to it.
$\vec{a} = \dfrac{qvB}{m}(-\sin 60^\circ\hat{i} - \cos 60^\circ\hat{j}) = \dfrac{qvB}{m}\left(-\dfrac{\sqrt{3}}{2}\hat{i} - \dfrac{1}{2}\hat{j}\right)$
So the answer is (a).

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