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Q.

A uniform magnetic field of magnitude 1 T exists in region $y \geq 0 along \hat{k}$ direction as shown. A particle of charge 1 C is projected from point $(- \sqrt{3}, - 1)$ towards origin with speed 1 m/s. If mass of particle is 1 kg, then co-ordinates of centre of circle in which particle moves are

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a

$(1, \sqrt{3},)$

b

$(1, - \sqrt{3},)$

c

$(\frac{1}{2}, - \frac{\sqrt{3}}{2})$

d

$(\frac{\sqrt{3}}{2}, - \frac{1}{2})$

answer is C.

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Detailed Solution

detailed_solution_thumbnail

The centre will be at 'C' as shown:
Coordinate of the centre are

$(r \cos 60^{0}, - r \sin 60^{0})$

$where r = radius of circle = \frac{mv}{Bq} = \frac{1 \times 1}{1 \times 1} = 1$

Question Image

$i . e ., (\frac{1}{2}, - \frac{\sqrt{3}}{2})$

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A uniform magnetic field of magnitude 1 T exists in region y≥ 0 along k^ direction as shown. A particle of charge 1 C is projected from point −3, −1 towards origin with speed 1 m/s. If mass of particle is 1 kg, then co-ordinates of centre of circle in which particle moves are