# Induced current from moving coil through non-uniform magnetic field

A rectangular coil with dimensions $L_x$ and $W_y$ , mass $m$, $n$ turns , and resistance $R$ is moving at an initial velocity of $v_i$ when a non-uniform magnetic field pointing directly up (perpendicular to the motion of the coil) is turned on. The magnetic field varies along the x-axis and is described by $B(t) = b_1x+b_2$.

1. What is the induced current the moment the field is turned on?
2. What is the initial deceleration of the coil?

Edit: 1. can be solved as follows: $I = (NLWb_1v_i)/R$

What's the best way to approach 2.?

• Use the fact that the force on one length of wire is given by: $$\vec F =\vec I \times \vec B$$ for each side then use newton's second law May 16 '15 at 7:27
• Isn't length part of that equation as well? $\vec F =\vec IL \times \vec B$
– WHY
May 16 '15 at 7:39
• How do I determine the direction of the force vector for each length of wire? Or are they all pointing in the $-x$ direction?
– WHY
May 16 '15 at 7:41
• @Joseph, your method worked. Thank you. If you'd like to post an official answer, I'd be happy to accept it.
– WHY
May 16 '15 at 7:56

The force on each side of the wire is given by: $$\vec F=L \vec I \times \vec B$$ (you where right about the length), and the direction of the force is given by the direction of the cross product.
If you want to be more general you can right the force as $$d \vec F=I d\vec l\times \vec B$$ Where $d \vec l$ is a small element of wire. And then integrate over the loop.