Question

A small, circular ring is inside a larger loop that is connected to a battery and a switch (Fig. E29.21). Use Lenz’s law to find the direction of the current induced in the small ring (a) just after switch S is closed; (b) after S has been closed a long time; (c) just after S has been reopened after it was closed for a long time.

Short answer

1. The direction of the induced current in the smaller loop is clockwise.
2. There is no induced current in the smaller loop.
3. The direction of the induced current in the smaller loop is counterclockwise.

Step-by-step solution

About Lenz law.

Lenz's law: Lenz's law states that an induced current or emf always tends to oppose or cancel out the change that caused it.

Right-hand rule for a magnetic field produced by a current in a loop: When the fingers of your right hand curl in the direction of the current, your right thumb points in the direction of the magnetic field lines.

Conditions on Lenz law

When the switch S is closed, the current flows in the circuit in the direction of the +

terminal to the - terminal of the battery.

Using the right-hand rule on the larger loop, we curl the fingers of the right hand, and

we find that the right thumb is pointing outward (.)(out of the screen)

Thus, whenever the switch S is closed, the magnetic field produced by the current in

the larger loop at the smaller loop is directed outward (.).

When the switch is closed.

(a) When switch S is closed after having been opened for a while, the current in the

the loop is increased from zero to a finite value, and so the magnetic field.

So, the magnetic field at the smaller loop is increasing.

Applying Lenz's law, this change in the magnetic field at the smaller loop induces a

current in the smaller loop to oppose this change.

So, the magnetic field produced by the induced current is to decrease the magnetic

field directed outward (.).

Thus, the magnetic field produced by the smaller loop is directed inward (x)(into the

screen)

Using the right-hand rule, point your right thumb in the direction of the magnetic field

inward (x), so your right fingers now curl in the direction of the induced current

Therefore, the direction of the induced current in the smaller loop is clockwise.

The switch is closed for a long time.

(b)When switch S is closed for a long time, the current in the loop does not change, and so does the magnetic field.

So, the magnetic field at the smaller loop is constant.

Since

$\Phi =cons\tan t,d\Phi /dt=0$

Applying Faraday’s law, the induced current in the smaller loop is:

$\epsilon =\frac{d\Phi }{dt}=0$

Therefore, there is no induced current in the smaller loop.

Switch is opened after closing a long time a long time.

(c) When switch S is reopened after having been closed for a long time, the current in the loop decreases to zero, and so does the magnetic field.

So, the magnetic field at the smaller loop is decreasing

Applying Lenz's law, this change in the magnetic field at the smaller loop induces a

current in the smaller loop to oppose this change.

So, the magnetic field produced by the induced current is to increase the magnetic

field directed outward(.).

Thus, the magnetic field produced by the smaller loop is directed outward (•)( out of

the screen).

Using the right-hand rule, point your right thumb in the direction of the magnetic field

outward (.), so your right fingers now curl in the direction of the induced current.

Therefore, the direction of the induced current in the smaller loop is counterclockwise