An electron moves in a circular orbit of radius 0.51A around a nucleus at a frequency of 6.8times10^(15)Hz .Estimate the magnetic field at the nucleus

Asked by bogarpradnya | 3rd Mar, 2020, 08:42: AM

Expert Answer:

Frequency, nu = 6.8 × 1015 Hz
Current, I = enu = 1.6 x 10-19 x 6.8 × 1015 =10.88 × 10-4 A =1.08 × 10-3 A
Magnetic field at centre is, 
 
B equals space fraction numerator mu subscript 0 i over denominator 2 a end fraction space equals space fraction numerator 4 pi cross times 10 to the power of negative 7 end exponent cross times 1.08 cross times 10 to the power of negative 3 end exponent over denominator 2 cross times 0.51 cross times 10 to the power of negative 10 end exponent end fraction space equals space 13.5 space T e s l a 
Thus,  B = 13.5 T 
 
 

Answered by Shiwani Sawant | 3rd Mar, 2020, 12:11: PM