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An electron moving with a speed of 10^(8...

An electron moving with a speed of `10^(8) ms^(-1)` enters a magnetic field of `5 xx 10^(-3)T` in a direction of perpendicular to field. Calculate
radius of the path

Text Solution

Verified by Experts

`evB = (mv^(2))/(r )`
`r= (mv)/(eB)` …(i)
`m= 9 xx 10^(-31)kg`
`v= 10^(8) ms^(-1)`
`e= 1.6 xx 10^(-19)C`
`B=5 xx 10^(-3)T`
radius `r= (9xx 10^(-31) xx 10^(8))/(1.6 xx 10^(-19) xx 5 xx 10^(-3))`
= 0.113m
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