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Find the flux density of the magnetic fi...

Find the flux density of the magnetic field to cause `62.5 eV ` electron to move in a circular path of radius 5 cm. Given `m = 9.1 xx 10^-31 kg, e = 1.6 xx 10^-19 C`

Text Solution

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`E = 62*5eV = 62*5 xx 1*6 xx 10^(-19)J`
Radius ` = r = 5 cm = 5 xx 10^(-2)` m,
`q = e = 1*6 xx 10^(-19)` C
`m = 9*1 xx 10^(-31)` kg
`E = (1)/(2) (B^(2)q^(2)r^(2))/(m)`
`B^(2) (2mE)/(q^(2)r^(2))/(q^(2)r^(2))`
`B = sqrt((2mE)/(q^(2)r^(2)))=(sqrt(2mE))/(qr)`
`B = sqrt(2xx9*1 xx 10^(-31) xx62*5 xx 10^(-19))/(1*6xx10^(-19)xx5xx10^(-2))`
`=(sqrt(1820xx10^(-50)))/(8*0xx10^(-21))`
` = (42*66 xx 10^(-25))/(8*0 xx 10^(-21))`
`rArr B = 5*33 xx 10^(-4)` T
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