Angular momentum due to electromagnetic induction

[Pushoam]

Homework Statement

Homework Equations

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?

 

[kuruman]

Magnetic force is along $\hat{s}$ direction.

Where does the magnetic force come from?

 

[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?

 

[haruspex]

option (d).
Is this correct?

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

 

[Pushoam]

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

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

 

[haruspex]

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

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.