# Homework Help: Angular momentum due to electromagnetic induction

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1. Dec 27, 2017

### Pushoam

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

2. Relevant equations

3. The attempt at a solution

$\frac { - d \phi }{dt} = V$

V denotes emf.

The current is in $\hat \phi$ direction.

Magnetic force is along $~\hat s$ direction.

Where $~\hat s$ is the radially outward direction in cylindrical coordinate system.

So, torque $\vec \tau$ about an axis passing through the center and perpendicular to the plane of loop is 0.

So, there is no change in angular momentum.

Hence, the option (b) is answer.

Is this correct?

2. Dec 27, 2017

### kuruman

Where does the magnetic force come from?

3. Dec 28, 2017

### Pushoam

There is no magnetic force.

Due to the change in flux, there is induced electric field.

Now, the induced current is in anti – clockwise direction according to Lenz's law.

So, the induced electric field should also be in anti – clockwise direction.

Force due to this induced electric field is $\vec F = \int_{ 0}^{ 2 \pi R} \vec E \lambda dl$ .....(1)

Due to the symmetry of the problem, $\vec E$ could be taken outside the integration.

$\vec F = { 2 \pi R} \vec E \lambda$ .....(2)

Torque about an axis passing through the center of the loop and perpendicular to the loop is $\tau = \vec R \times \vec F$ .....(3)

Change in the angular momentum , $\Delta \vec L = R ~ { 2 \pi R} E \lambda ~dt ~\hat z$ .....(4)

Now, $d \phi = - B \pi a^2 = - V dt = - \int_{0 }^{ 2 \pi R } \vec E . d\vec l$ dt .....(5)

Due to the symmetry of the problem, $\int_{0 }^{ 2 \pi R } \vec E . d\vec l = { 2 \pi R} E$ .....(6)

So, ${ 2 \pi R} ~E dt = B \pi a^2$ .....(7)

From (4) and (7),

$\Delta \vec L = R ~ B \pi a^2\lambda ~\hat z$ .....(8)

So, the answer is $\Delta L = \pi a^2 RB \lambda$ , option (d).

Is this correct?

4. Dec 28, 2017

### haruspex

Options b) and d) are the only two that make sense dimensionally.

5. Dec 28, 2017

### Pushoam

Among (b) and (d), the answer is (d).
Right?

6. Dec 28, 2017

### haruspex

I would say so.
The diagram threw me, though. It makes it look as though the field lines are parallel to the plane containing the ring. The verbal description implies they're normal to it.