Asked by  | 3rd Sep, 2009, 04:40: PM

Expert Answer:

F = q(v x B)

Proton's position vector = aj

Proton's initial velocity = vk

The magnetic field at proton's location, B = Bk = μ0ik/(2πa)

{Curl the right hand fingers around the conductor with thumb pointing in the direction of +x i.e. direction of current.

The curled fingers represent magnetic field lines, the tangent at proton's location points in the +z direction}

F = q(v x B) = +e(vk x Bk) = 0

Electron's position vector = -aj

Electron's initial velocity = vi 

The magnetic field at proton's location, B = B(-k) = μ0i(-k)/(2πa)

F = q(v x B) = -e(vi x B(-k))

= -evBj

= -evμ0i j/(2πa)

The magnetic force on the proton is zero, while electron experiences a magnetic force in -y direction.




Answered by  | 3rd Sep, 2009, 10:33: PM

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