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CBSE Class 12-science Answered

A proton is moving around the equator of the earth with the speed of begin mathsize 11px style 2 cross times 10 to the power of 5 space straight m divided by straight s end style. Find the minimum magnetic field, which should be created at the equator for this purpose. The mass of proton begin mathsize 12px style 1.7 cross times 10 to the power of negative 27 end exponent kg end style  and the radius of earth = begin mathsize 12px style 6.37 cross times 10 to the power of 6 straight m end style.
Asked by Topperlearning User | 20 May, 2015, 09:53: AM
answered-by-expert Expert Answer

Given,

v = begin mathsize 11px style 2 cross times 10 to the power of 5 space straight m divided by straight s end style

m = begin mathsize 12px style 1.7 cross times 10 to the power of negative 27 end exponent kg end style

r = begin mathsize 12px style 6.37 cross times 10 to the power of 6 straight m end style

q = 1.6 x 10-19 C

The magnetic field is given as,

begin mathsize 11px style straight B equals mv over qr equals fraction numerator 1.7 cross times 10 to the power of negative 27 end exponent cross times 2 cross times 10 to the power of 5 over denominator 1.6 cross times 10 to the power of negative 19 end exponent cross times 6.37 cross times 10 to the power of 6 end fraction equals 0.335 cross times 10 to the power of negative 9 end exponent straight T end style

Answered by | 20 May, 2015, 11:53: AM

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