Force Between 2 Coaxial Current-Carrying Loops

[KittyKinkle02]

Homework Statement

A flat coil having 50 turns of radius 30cm is in series with another flat coil of 100 turns each of radius 2cm. The coils are coaxial and 10cm apart.

Estimate the force between the coils when the current passing is 2A.

Relevant Equations

Biot Savart Law: $\vec{dB} = \frac{\vec{dl} \times \hat{r} \mu_0 I}{4\pi r^2}$
Force on a current carrying conductor: $F = BIL$

I have attempted to solve it as follows:

Using the Biot-Savart law, I found the flux density at the centre of the smaller coil due to the bigger coil as:
$$\frac{\mu_0 I b^2 N_2}{2(a^2 + b^2)^{1.5}}$$
where a is the distance to the coil (10cm), N2 is the number of loops in the larger coil (50), and b is the radius of larger coil (30cm).

I found the value of magnetic flux density at the centre of the smaller coil to be approximately 1.78 x 10^-4 T.

The textbook's answer section says to assume that this constant over the smaller coil. So using $F = BIL$ I get F = 1.78 x 10^-4 T * 2 * 2*pi*(0.02)*100 = 4.49 x 10^-3 N. The textbook says that the answer is 1.4 x 10^-4 N (a factor of ~32 from my answer).

I feel like the last step with F=BIL is wrong (Have I done the line integral incorrectly?).

 

[kuruman]

I feel like the last step with F=BIL is wrong

You are using the Lorentz force expression on a wire segment of length $L$. It might work if you know what you're doing. You don't show the details.
I would assume that the smaller coil is a point dipole in the magnetic field of the larger coil.

1. What is its magnetic dipole moment?
2. What is the magnetic potential energy of the configuration?
3. How can you get a force from the potential energy?

 

[KittyKinkle02]

The magnetic dipole moment should be $\vec{m} = N_1 I \vec{A}$ with $\vec{A}$ being a vector normal to the surface bound by the current loop.

The potential energy is given by $U = -\vec{m} \cdot \vec{B}$ but since m and B point in the same direction it should just be $U = -mB$.

The potential energy is thus $N_1 I A B$ = 100 * 2 * pi*(0.02)^2 * 1.78x10^-4 = 4.47x10^-5 J.

But I don't see how this leads me any closer to the solution?



For the Lorentz force part I used: $$\vec{F} = \oint_{loop} I \vec{dl} \times \vec{B_{ext}}$$

The principle of superposition means that I should be able to multiply it by $N_1$ and get the total force.

The B field is constant around the loop and always perpendicular to dl, so I simplified the integral to this (r is the radius of the smaller coil): $$N_1 \int_0^{2\pi r} BI dl$$

Which becomes B * I * 2pi * r * N_1 ~= 4.5x10^-3.

 

[kuruman]

But I don't see how this leads me any closer to the solution?

You didn't answer question 3 to post #2. Just to get you thinking in the right direction.
The gravitational potential energy of the Earth-Moon system is a scalar field given by $$U=-\frac{GM_eM_m}{r}.$$ The Moon is at distance $R_m$ from the Earth. How do you get the force exerted by the Earth on the Moon (as expressed by Newton's law of gravitation) given the above information?

 

[kuruman]

For the Lorentz force part I used: $$\vec{F} = \oint_{loop} I \vec{dl} \times \vec{B_{ext}}$$

The principle of superposition means that I should be able to multiply it by $N_1$ and get the total force.

The B field is constant around the loop and always perpendicular to dl, so I simplified the integral to this (r is the radius of the smaller coil): $$N_1 \int_0^{2\pi r} BI dl$$

Which becomes B * I * 2pi * r * N_1 ~= 4.5x10^-3.

Not how it works. If you have a current circuit in a uniform magnetic field the net force it is zero. That's because if the field is uniform, one can take it and the current out of the integral $$\vec{F} = I\left(\oint_{loop} \vec{dl}\right) \times \vec{B}_{ext}$$in which case the term in parentheses is zero. It's easier to see this if you consider the net Lorentz force on a rectangular current loop placed in a uniform magnetic field where the forces cancel in pairs.