Question

What should be the minimum magnitude of the magnetic field that must be produced at the equator of earth so that a proton may go round the earth with a speed of 1 × 10^7 ms^-1? Earth’s radius is 6.4 × 10^6 m.

(a) 16.63 × 10^-8 T

(b) 91.63 × 10^-8 T

(c) 1.63 × 10^-8 T

(d) 761.63 × 10^-8 T

The question was posed to me in quiz.

This key question is from Motion in Combined Electric and Magnetic Field in division Moving Charges and Magnetism of Physics – Class 12

Answer 1

The correct answer is (c) 1.63 × 10^-8 T

For explanation I would say: B = \(\frac {mv}{qr}\).

B = \(\frac {(1.67 \, \times \, 10^{-27} \, \times \, 10^7)}{(1.6 \, \times \, 10^{-19} \, \times \, 6.4 \, \times \, 10^6)}\)

B = 1.63 × 10^-8 T

Therefore, the minimum magnitude of magnetic field should be 1.63 × 10^-8 T.