an electron having a K.E of 100eVcirculates in a path of radius 10cm in a magnetic field.find the magnetic field and the no of revolution per second made by the electron

Dear Student,

Given that K.E of electron:

K=12mv2=100 eVor 12m2v2m=100 eVor mv=2m×100 eVwhere, m is mass of electron
Now, we know that radius, 'r' of an electron in an external applied magnetic field  'B' is :

r=mvqBwhere, vis velocity of electron ; q is its charge.Therefore, B=mvqrgiven r=10 cm=0.1 m.and we know, q=1.6×10-19 C and m=9×10-31 kg, for an electron.plugging in these values along with the value of mv, we get:B=2×9×10-31×100×1.6×10-191.6×10-19×0.1    =3.35×10-4 T=3.35 Gauss.

and, no. of revolutions made by electron per second is given by:

n=qB2πm=1.6×10-19×3.35×10-42×3.14×9×10-31n=9.4×106 revsec.

Regards,
Manoj Singh.
 

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