Question

An electron that has a velocity with x component 2.0 × 106 m/s and y component 3.9 × 106 m/s moves through a uniform magnetic field with x component 0.035 T and y component -0.21 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Answer 1

magnetic force Fb = q*(v X B) = q*( V cross B)

Fb = -1.6*10^-19*( (2*10^6 i + 3.9*10^6 j) X (0.035i - 0.21j) )

Fb = (1.6*10^-19*2*10^6*0.21) k + (1.6*10^-19*3.9*10^6*0.035) k

Fb = 8.904*10^-14 N <<<---ANSWER

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for proton

magnetic force Fb = q*(v X B) = q*( V cross B)

Fb = 1.6*10^-19*( (2*10^6 i + 3.9*10^6 j) X (0.035i - 0.21j) )

Fb = -(1.6*10^-19*2*10^6*0.21) k - (1.6*10^-19*3.9*10^6*0.035) k

Fb = -8.904*10^-14 N <<<---ANSWER