A loop of radius a and mass M made of resistive material
has a total resistance R. At t = 0, the loop is located at x = 0 and moves with
velocity v_{0 }i. The loop lies in the x-y plane. There is a
magnetic field **B** = B_{0} (x/x_{0})
**k**. How far
does the loop travel before stopping?

Solution:

- Concepts:

Faraday's law - Reasoning:

The magnetic flux through a filamentary loop is changing with time. This flux change induces an emf. The emf causes a current to flow, which dissipates energy. - Details of the calculation:

Flux F = πa^{2}B = πa^{2}B_{0}x/x_{0}. |emf| = ∂F/∂t = (πa^{2}B_{0}/x_{0})∂x/∂t = (πa^{2}B_{0}/x_{0})v.

I = emf/R. P = I*emf = emf^{2}/R. The power dissipated must equal the rate at which the loop looses kinetic energy.

emf^{2}/R = -Mvd(v/dt). dv/dt = -bv with b = (πa^{2}B_{0}/x_{0})^{2}/(RM).

v(t) = v_{0}exp(-bt).

The distance the loop travels before stopping is

d = ∫_{0}^{∞}v(t)dt = -(1/b)v_{0}exp(-bt)|_{0}^{∞}= v_{0}/b = v_{0}RMx_{0}^{2}/(πa^{2}B_{0})^{2}.

(a) A closed circular coil of N turns, radius a and total resistance R
is rotated with uniform angular velocity ω about a vertical diameter in a
horizontal magnetic field **B**_{0} = B_{0}**i**.
Compute the emf ε induced in the coil, and also the mean power
<P> required for maintaining the coil's motion. Neglect the coil self
inductance.

(b) A small magnetic needle is placed at the center of the coil, as shown in the
figure. It is free to turn slowly around the z-axis in a horizontal plane, but
it cannot follow the rapid rotation of the coil.

Once the stationary regime is reached, the needle will point in a direction
making a small angle θ with **B**_{0}. Compute the resistance R of the coil in terms of
this angle and the other parameters of the system. (Lord Kelvin used this method in the 1860s to set the absolute standard for
the ohm.)

Solution:

- Concepts:

Induced emf, the magnetic field at the center of a current loop - Reasoning:

The changing magnetic flux through the rotating loop produces an emf, which causes a current to flow. This current produces its own magnetic field. - Details of the calculation:

(a) After some time t, the normal to the coil plane makes an angle ωt with the magnetic field**B**_{0}= B_{0}**i**. Then, the magnetic flux through the coil is Φ = N**B**∙**S**, where the vector surface**S**is given by

**S**= πa^{2}(cos(ωt)**i**+ sin(ωt)**j**). Therefore Φ = NB_{0}πa^{2}cos(ωt).

The induced emf is ε = -dΦ/dt = NB_{0}πa^{2}ω sin(ωt).

The instantaneous power is P = ε^{2}/R. The mean power is <P> = (NB_{0}πa^{2}ω )^{2}/(2R).

(b) The total field at the center the coil at the instant t is**B**_{t}=**B**_{0}+**B**_{i}, where**B**_{i}is the magnetic field due to the induced current

B_{i}= B_{i}(cos(ωt)**i**+ sin(ωt)**j**) with B_{i }= μ_{0}NI/(2a) and I = ε/ R.

[cos(ωt)**i**+ sin(ωt)**j**is the direction of the vector surface**S**.]

Therefore B_{i}= μ_{0}N^{2}B_{0}πaω sin(ωt)/(2R).

The mean values of its components are

<B_{ix}> = [μ_{0}N^{2}B_{0}πaω/(2R)]<sin(ωt) cos(ωt)> = 0,

<B_{iy}> = [μ_{0}N^{2}B_{0}πaω/(2R)]<sin^{2}(ωt)> = μ_{0}N^{2}B_{0}πaω/(4R)].

For the total field we have

<**B**_{t}> = B_{0}**i**+ μ_{0}N^{2}B_{0}πaω /(4R)]**j**.

The needle orients along the mean field, therefore tanθ = μ_{0}N^{2}πaω/(4R).

Finally, the resistance of the coil measured by this procedure, in terms of is

R = μ_{0}N^{2}πaω/(4tanθ).

A square loop of wire with sides of length L carries a time-independent current I and is located at the origin. A rectangular loop of wire of length l and width W is moving parallel to the x-axis in the positive direction at a constant velocity v. Its width is so small that the variation of the field along its width may be neglected. At t = 0, the nearest edge of the rectangular loop has the coordinates (a,0,a), where a >> L. Find the induced emf in the rectangular loop at t = 0.

Solution:

- Concepts:

Dipole moment, dipole field, induced emf - Reasoning:

The small current loop at the origin has a dipole moment, and its magnetic field is approximately a dipole field. The rate of change of the flux of this field through the rectangular loop is proportional to the induced emf in the rectangular loop. - Details of the calculation:

Let us use Gaussian units.

The dipole moment of the loop is**m**=**k**IA =**k**IL^{2}= k m.

The magnetic field produced by this loop at r (r >> L) is

**B**= k_{m}(3(**m∙r**)**r**-**m**r^{2})/r^{5}= k_{m}(3mz**r**-**k**mr^{2})/r^{5}. (k_{m}= μ_{0}/(4π))

The induce emf is ε = -(∂F/∂t), with F = ∫_{A}(B∙k)dA.

**B∙k**= k_{m}(3mz^{2}- mr^{2})/r^{5}= k_{m}m(2z^{2}- x^{2})/(x^{2}+ z^{2})^{5/2}.

At t = 0 F = ∫_{A}(**B∙k**)dA = Wk_{m}∫_{a}^{a+l}dx m(2a^{2}- x^{2})/(a^{2}+ x^{2})^{5/2 }After the loop has moved a small distance dx', the flux is

F = Wk_{m}∫_{a+dx'}^{a+l+dx'}dx m(2a^{2}- x^{2})/(a^{2}+ x^{2})^{5/2 }The change in flux therefore is

∂F/∂x' = -Wk_{m}m(2a^{2}- a^{2})/(a^{2}+ a^{2})^{5/2}+ Wk_{m}m(2a^{2}- (a + l)^{2})/(a^{2}+ (a + l)^{2})^{5/2 }= Wk_{m}m[-1/(2^{5/2}a^{3}) + a^{2}(1- (l^{2}+ 2al)/a^{2})/(2a^{2}(1 + (l^{2}+ 2al)/(2a^{2})))^{5/2}]

= Wk_{m}m(1/(2^{5/2}a^{3}))[-1 + (1 - (l^{2}+ 2al)/a^{2})/(1 + (l^{2}+ 2al)/(2a^{2}))^{5/2}]

∂F/∂t = (∂F/∂x')(∂x'/∂t)

= Wk_{m}mv(1/(2^{5/2}a^{3}))[-1 + (1 - (l^{2}+ 2al)/a^{2})/(1 + (l^{2}+ 2al)/(2a^{2}))^{5/2}]

ε = -(∂F/∂t) = Wk_{m}mv(1/(2^{5/2}a^{3}))[1 - (1 - (l^{2}+ 2al)/a^{2})/(1 + (l^{2}+ 2al)/(2a^{2}))^{5/2}]