An ion beam of specific charge 5 × 10^7 coulomb/kg enter a uniform magnetic field of 4 × 10^{-2},

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4.Moving Charges and Magnetism

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An ion beam of specific charge $5 \times 10^7$ $coulomb/kg$ enter a uniform magnetic field of $4 \times 10^{-2}\, tesla$ with a velocity $2 \times 10^5\, m/s$ perpendicularly. The radius of the circular path of ions in meter will be

A

$0.10$

B

$0.16$

C

$0.20$

D

$0.25$

Solution

Given $: \frac{e}{m}=5 \times 10^{7} \mathrm{C} / \mathrm{kg}$

In a cyclotron, radius of the circular path,

$r=\frac{m v}{B e}=\frac{2 \times 10^{5}}{5 \times 10^{7} \times 4 \times 10^{-2}}=0.1 \mathrm{m}$

Standard 12

Physics

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