Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days


Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number


Mon to Sat - 11 AM to 8 PM

CBSE - XII Science - Physics - Moving Charges and Magnetism


Asked by 3rd September 2009, 4:40 PM
Answered by Expert


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 Expert 3rd September 2009, 10:33 PM

Rate this answer

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Report an issue
Your answer has been posted successfully!

Related Question

Answered by Expert
Answered by Expert
Answered by Expert
Answered by Expert

Latest Questions