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CBSE Class 12-science Answered

The electron in a hydrogen atom circles around the proton with a speed of 2.18 x 106 m/s in an orbit of radius 5.3 x 10 -11 m. calculate the equivalent current and magnetic field produced by proton.
Asked by Topperlearning User | 18 May, 2015, 04:14: PM
answered-by-expert Expert Answer

v = 2.18 x 106 m/s

r = 5.3 x 10 -11 m

Time period (T)= begin mathsize 11px style fraction numerator 2 πr over denominator straight nu end fraction end style = begin mathsize 11px style fraction numerator 2 cross times 3.14 cross times 5.3 cross times 10 to the power of negative 11 end exponent over denominator 2.18 cross times 10 to the power of 6 end fraction space equals space 1.528 cross times 10 to the power of negative 16 end exponent space straight s end style

 

Equivalent current I = e/T

 

begin mathsize 11px style straight I equals fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent over denominator 1.528 cross times 10 to the power of negative 16 end exponent end fraction equals 1.048 cross times 10 to the power of negative 3 end exponent space straight A end style

 

Magnetic field produced by proton

begin mathsize 11px style straight B equals fraction numerator straight mu subscript ring operator over denominator 4 straight pi end fraction fraction numerator 2 πI over denominator straight r end fraction space equals space fraction numerator 10 to the power of negative 7 end exponent cross times 2 cross times 3.14 cross times 1.048 cross times 10 to the power of negative 3 end exponent over denominator 5.3 cross times 10 to the power of negative 11 end exponent end fraction end style

 B = 12.42T

 

Answered by | 18 May, 2015, 06:14: PM
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