MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics 8.02
Force and Torque on a Current Loop in a Uniform Magnetic Field
Solutions
Problem: Consider the current carrying loop at right which
sits in a uniform external magnetic field B. What is the force
and torque on the loop? What motion does it make?
Solution:
1) What is the net force?
The forces on the top and bottom (legs 1 & 3) of the loop are
both zero, as the current is parallel to the magnetic field. The
forces on legs 2 & 4 are equal and opposite and hence cancel. So the net magnetic force is zero.
2) What is the net torque?
The force on leg 2 is out of the page and the force on
leg 4 is into the page, so there will be a torque up
(which means that the loop wants to rotate into the
page on the right). To calculate its magnitude we need
to arbitrarily choose an axis of rotation. I’ll pick leg 2.
Then the torque only comes from leg 4 and we have:
  
τ = r × F = bî × IaB −k̂ = IabBĵ
( )
We can show this in more detail. We shall calculate the torque about the origin.

Consider a infinitesimal current element on leg 4, given by Id s = Idyĵ , shown in the figure above.

The force on this element is dF4 = (Idyĵ × Bî) = (− IdyBk̂) . The torque on this element is

 

d τ 4 = r0,4 × dF4 . The vector r0,4 = (bî + yĵ) . Therefore the torque is
p. 1 of 2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics 8.02


d τ 4 = (bî + yĵ) × dF4 = (bî + yĵ) × (Idyĵ × Bî)
= (bî + yĵ) × (− IdyBk̂) = IbdyBĵ − IydyBî
To find the torque on leg 4 we need to integrate

τ4 =
y=+ a/2
∫
y=− a/2

d τ4 =
y=+ a/2
1
(IbdyBĵ − Iy dyBî) = IbaBĵ − IB y 2
∫
2
y=− a/2
y=a/2
y=− a/2
î = IbaBĵ .


A similar calculation for leg 1 leads to zero because r2 = yĵ , dF2 = (− Idyĵ × Bî) = IdyBk̂ . Hence

d τ 2 = ( yĵ) × (IdyBk̂) = IydyBî . This is zero after integration

τ2 =
y=+ a/2
1
Iy dyBî = IB y 2
2
y=− a/2
∫
y=a/2
y=− a/2

î = 0 .
The torques on legs 1 and 3 will add to zero because the force is in zero on those legs. Therefore
 
the torque on the loop is τ = τ 4 = IbaBĵ .
So what happens to the loop due to this torque? It will rotate until it has rotated 90 degrees. At
this point the forces on legs 2 and 4 will be “outward,” making the loop want to expand rather
than rotate. So at this point the loop won’t want to rotate any more – it is an equilibrium position.
Of course the loop will already have angular momentum from the motion it just underwent
meaning that it will continue past this equilibrium point and then undergo harmonic motion
about it.
p. 2 of 2