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Question: A proton an \(\alpha\)-particle enter in a uniform magnetic field perpendicularly with same speed. T...

A proton an α\alpha-particle enter in a uniform magnetic field perpendicularly with same speed. The ratio of time periods of both particle (TpTα)\left( \frac { T _ { p } } { T _ { \alpha } } \right) will be

A

1 : 2

B

1 : 3

C

2 :1

D

3 : 1

Answer

1 : 2

Explanation

Solution

The time period of revolution of a charge particle in a magnetic filed is

For proton,

Now, for α−\alpha -particle

mα=4 m,qα=2q\mathrm { m } _ { \alpha } = 4 \mathrm {~m} , \mathrm { q } _ { \alpha } = 2 \mathrm { q }

∴Tα=2π(4 m)B(2q)=2(2πmBq)⇒TpTα=12\therefore \mathrm { T } _ { \alpha } = \frac { 2 \pi ( 4 \mathrm {~m} ) } { \mathrm { B } ( 2 \mathrm { q } ) } = 2 \left( \frac { 2 \pi \mathrm { m } } { \mathrm { Bq } } \right) \Rightarrow \frac { \mathrm { T } _ { \mathrm { p } } } { \mathrm { T } _ { \alpha } } = \frac { 1 } { 2 }