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Question
A long straight wire AB carries a current of 5A. P is a proton travelling with a velocity of 2 × 106 m/s, parallel to the wire, 0.2 m from it and in a direction opposite to the current, as shown in Figure below. Calculate the force which magnetic field of the current carrying conductor AB exerts on the proton.
Solution
The magnetic field due to the current carrying wire is perpendicular to the plane of the paper, in a downward direction.
i.e., `vec"B" = - (mu_0"I")/(2pi"d") vec"k"`
Force `vec "F" = q vecv xx vec"B"`
`= e(- v vec"j") xx (- (mu_0"I")/(2pi"d") vec"k")`
`= (mu_0ev"I")/(2pi"d") vec "i"`
Given that d = 0.2 m, ν = 2 × 106 m/s, I = 5 A
`therefore "F" = (mu_0"e" xx 2 xx 10^6 xx 5 xx 10^5)/(2pi xx 0.2) vec"i"`
`= (2 xx 10^-7 xx 1.6 xx 10^-19 xx 2 xx 10^6 xx 5 xx 10^5)/0.2`
`= 160 xx 10^-13` N
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