# Homework Help: Torque on a Coil

1. Aug 4, 2014

### Zondrina

1. The problem statement, all variables and given/known data

The attachment shows a rectangular $20-turn$ coil of wire, of dimensions $10 cm$ by $5.0 cm$. It carries a current of $0.10A$ and is hinged along one long side. It is mounted in the x-y plane, at angle $θ = 30°$ to the direction of a uniform magnetic field of magnitude $0.50 T$. In unit-vector notation, what is the torque acting on the coil about the hinge line?

2. Relevant equations

$(1) \quad \vec F_B = i(\vec L \times \vec B)$, where $\vec L$ is assumed in the direction of conventional current.

$(2) \quad \vec{\tau} = \vec{\mu} \times \vec B$, where $\vec{\mu}$ is the magnetic dipole moment with magnitude $NiA$.

3. The attempt at a solution

The magnetic field has components given by:

$\vec B_x = B cos(\theta) \hat i = (0.50 T) cos(30°) \hat i = (0.433 \hat i) T = (0.43 \hat i) T$
$\vec B_z = B sin(\theta) \hat j = (0.50 T) sin(30°) \hat j = (0.25 \hat j) T$

For the portion of coil along the torque axis, the $x$ component passes through the axis and therefore causes no torque about it.

For the two horizontal portions of coil, the angle between the magnetic field and length vector is zero, so that $\vec F_B = 0$. Hence no torque.

The $x$ component causes torque only for the portion of the coil located at $x = 0.050 m$ and nowhere else along the coil. This is because the direction of the magnetic force given by the right hand rule is in the $+ z$ direction for that segment of coil.

The $z$ component passes through the hinge line. Therefore, it causes no torque about the axis for the segment of coil along the axis.

Examining the two horizontal portions of the coil, we can observe no torque due to the $z$ component again. This is because if the fingers on the right hand are extended in the direction of the conventional current, the right hand rule gives two magnetic forces pointing in the $+y$ and $-y$ directions. These forces are along the axis of rotation, and therefore produce no torque about it.

For the segment of coil located at $x = 0.050 m$, the $z$ component again causes no torque. This is because the right hand rule gives the magnetic force in the $- x$ direction. The force passes through the torque axis, hence no torque about that axis.

So the $z$ component causes no net torque at all. The net torque seems to be caused by only the $x$ component of the magnetic field acting along the length of coil at $x = 0.050 m$.

Assuming counter-clockwise is positive, the magnetic force acting on that part of the coil is given by:

$\vec F_{B_x} = i(\vec L \times \vec B_x) = (0.10 A)[(- 0.10 m)\hat k \times (0.433 T) \hat i] = (0.00433 \hat j) N = (0.0043 \hat j) N$

Hence the force points in the $+ z$ direction. The torque caused by this force can be found by using a position vector $\vec r = (0.050 \hat i) m$:

$\vec{\tau} = \vec r \times \vec F_{B_x} = ((0.050 \hat i)m) \times ((0.00433 \hat j) N) = (- 0.00022 \hat k) N \cdot m$

Since counter-clockwise was assumed positive and we got a negative answer, the torque is actually in the clockwise direction about the hinge axis.

Could we have used equation $(2)$?

$\vec{\tau} = \vec{\mu} \times \vec B_x = (20*0.10*0.0050 \hat j) \times (0.433 \hat i) N \cdot m = (0.00433 \hat k) N \cdot m = (0.0043 \hat k) N \cdot m$

Why are the answers different? How would I do this properly if I messed up something somewhere?

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Last edited: Aug 4, 2014
2. Aug 4, 2014

### TSny

Did you take into account the 20 turns?

Everything else looks very good to me.

3. Aug 4, 2014

### Zondrina

Ohhhh -facepalm- sorry about that. The magnetic force acts on each coil separately, and so:

$\vec F_{net} = 20 * \vec F_{B_x}$

Then $\vec{\tau} = \vec r \times \vec F_{net} = (0.0043 \hat k) N \cdot m$.

Thank you.

4. Aug 4, 2014

### Zondrina

Upon further inspection I noticed that when I calculated the torque using $(2)$:

$\vec{\tau} = \vec{\mu} \times \vec B_x = (20*0.10*0.0050 \hat j) \times (0.433 \hat i) N \cdot m = (0.00433 \hat k) N \cdot m = (0.0043 \hat k) N \cdot m$

I assumed that counter-clockwise was positive and got a positive answer. This should not be correct.

Applying the right hand rule properly this time $\vec{\mu}$ is actually in the $- \hat j$ direction. Then assuming counter-clockwise would produce a consistent result with the $\vec{\tau} = \vec r \times \vec F_{net}$ calculation.

Sorry for the formalities.

5. Aug 4, 2014

### TSny

Oops, I didn't notice the mistakes in unit vectors. The axis of rotation is along the y axis. Your torque vector should therefore be parallel to the y-axis (either +j or -j).

μ is not in the -j direction. Notice how the axes are labeled.

6. Aug 4, 2014

### Zondrina

Ah I was curious about that. I switched my vectors around and got the right answer though.

From now on I'll probably re-define the axes for simplicity.

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