A beam of ions with velocity 2 × {10^5},m/s enters normally into a uniform magnetic field of 4 × {

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4.Moving Charges and Magnetism

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A beam of ions with velocity $2 \times {10^5}\,m/s$ enters normally into a uniform magnetic field of $4 \times {10^{ - 2}}\,tesla$. If the specific charge of the ion is $5 \times {10^7}\,C/kg$, then the radius of the circular path described will be.......$m$

A

$0.10$

B

$0.16$

C

$0.20$

D

$0.25 $

Solution

(a) $r = \frac{{mv}}{{Bq}}$$ = \frac{v}{{(q/m)B}} = \frac{{2 \times {{10}^5}}}{{5 \times {{10}^7} \times 4 \times {{10}^{ – 2}}}} = 0.1\,m$

Standard 12

Physics

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