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A beam of ions with velocity 2xx10^(5)ms...

A beam of ions with velocity `2xx10^(5)ms^(-1)` enters normally into a uniform magnetic field of 0.04T. If the specific charge, i.e `(q)/(m)` of ion is `5xx10^(7)Ckg^(-1)`, find the radius of the circular path described.

Text Solution

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Since, the beam of ions enters normally into a uniform magnetic field, this means the path traversed by the ions will be circular in nature.
As, radius of circular path,
`r=(mvsintheta)/(qB)=(mvbotsin90^(@))/(qB)`
Given, `B=0.04T,v=2xx10^(5)m//s` and `(q)/(m)=5xx10^(-7)Ckg^(-1)`
`impliesr=(mv)/(qB)=(2xx10^(5))/(5xx10^(7)xx0.04)`
`=(2xx10^(5)xx10^(2))/(5xx10^(7)xx4)`
`=10^(-1)=0.1m`
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