1.

When alpha-particle and proton move perpendicular to uniform magnetic field with same speed then ratio of periodic times of their circular motion will be ______

Answer»

`2:1`
`1:2`
`4:1`
`1:4`

Solution :For UNIFORM circular motion,
`(MV^(2))/r=Bqv`
`therefore(mv)/r=Bq`
`therefore(m(romega))/r=Bq`
`thereforeomega=(Bq)/m`
`therefore(2pi)/T=(Bq)/m`
`thereforeT=(2pim)/(Bq)(because2piandB" are constants")`
`thereforeTpropm/q`
`thereforeT_(ALPHA)/T_(p)=m_(alpha)/m_(p)xxq_(p)/q_(alpha)`
`=(4m_(p))/m_(p)xxe/(2E)=2/1=2:1`


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