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Find the flux density of the magnetic fi...

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

Text Solution

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`E= 62.5 eV= 62.5 xx 1.6 xx 19^(-19)J`
Radius =r=5cm `=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))`
`B= sqrt((2mE)/(q^(2)r^(2)))= sqrt((2mE)/(qr))`
`B= (sqrt(2xx 9.1 xx 10^(-31) xx 62.5 xx 1.6 xx 10^(-19)))/(1.6 xx 10^(-19) xx 5 xx 10^(-2))`
`=sqrt((1820 xx 10^(-50))/(8.0 xx 10^(-21)))`
`=(42.66 xx 10^(-25))/(8.0 xx 10^(-21))1`
`B= 5.33 xx 10^(-4)T`
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