Question

A proton from cosmic rays Enter the Earth's magnetic field in a directionon a perpendiculace to the field of the velocity of proton is 2 x 107 m/s and B= 1.6x10-6 wb/m², find the force Exerted on the proton by the magnetic field (change on a proton (e) = 1.6 x 10-19 c)
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Answer 1

we know that
Force = qVB(sin 90= 1))
so after putting all the value
= 1.6*2*2*10⁻¹⁸N
5.12*10⁻¹⁸N
Answer