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Q.

A proton moving with a velocity $3 \times 10^{5} m / s$ enters a magnetic field of 0.3 tesla at an angle of 30º with the field. The radius of curvature of its path will be (e/m for proton $= 10^{8} C / kg)$

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a

1.25 cm

b

0.02 cm

c

0.5 cm

d

2 cm

answer is B.

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Detailed Solution

$r = \frac{mvsin θ}{Be} = \frac{3 \times 10^{5} \sin 30 °}{0.3 \times 10^{8}}$

$\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}} = 0.5 \times 10^{- 2} m = 0.5 cm$

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A proton moving with a velocity 3×105 m/s enters a magnetic field of 0.3 tesla at an angle of 30º with the field. The radius of curvature of its path will be (e/m for proton =108C/kg