Chapter 8.2, Problem 1aT

The resistance of loop 2 is greater than that loop l. (The loop are made from different materials.)

1. Is there a current induced through the wire of either of the loops:

• before the switch is closed? Explain.
• just after the switch is closed? Explain.
• a long after the switch is closed? Explain.

2. For the period of time that there is a current included through the wire of the loops, find the direction of the current.
3. The ratio of the induced currents for the two loops is found by experiment to be equal to the inverse of the ratio of the resistances of the loops.

What does this observation imply about the ratio of the induced emf in loop 1 to the induced emf in loop 2?

To determine

To Identify:

Induced current through wire of the loops as per the given conditions:

1. Before the switch is closed.
2. After when the switch is closed.
3. After a long-time when the switch is closed.

Explanation of Solution

Introduction:

According to Faradays’ law, an e.m.f is induced in a loop of wire if there is a rate of change in flux passing through the wire.

  $inducedemf=-\frac{d\varnothing }{dt}$

Where, $\varnothing$ is the flux.

Case1: Before the switch is closed:

Before the switch is closed, there is no current flowing in solenoid (bigger loop) that can produce changing magnetic field. Hence, there is no change in flux in the small loops. Therefore, there is no induced current in small coils.

Case 2: After the switch gets closed:

Just after the switch is closed, the current in the solenoid (bigger loop) goes from zero to maximum which makes the magnetic field lines passing through the small loops change. Hence, due to change in flux, there will be induced current in them. The loop with higher resistance will be associated with less induced current.

Case 3: After longtimewhen the switch isclosed:

After long time the switch is closed, there is constant current in solenoid (bigger loop) that produces constant magnetic field. Hence, there is no change in flux in the small loops. Therefore, there is no current in small loops.

Conclusion:

Therefore, following Faraday’s law, there is an induced current in small coils just when switch is closed and is zero for other cases.

To determine

To Find

The direction current induced through a wire of the loops.

Explanation of Solution

Introduction:

According to Lenz’s law, the induced emf will form a magnetic field which counteracts the change in flux.

By seeing the sign of the battery (current flows from positive to negative terminal) and using the right-hand rule, the direction of magnetic field induced in the greater loop must be from left to right. When switch is closed, then the induced current in small loop is in such a way that decreases the magnetic flux produced by the larger loop, hence, the induced current is in clockwise while seeing the loop from right. When switch is opened, the induced current will flow in anticlockwise direction.

Conclusion:

Therefore, the current induced through a wire of the loop will be such that it will oppose the change in flux produced by the bigger loop.

To determine

To Explain:

The ratio of induced emf in the loop 1to the loop 2.

Answer to Problem 1aT

Ratio of emf induced in loop 1to loop 2 is equal.

Explanation of Solution

Introduction:

According to Faradays’ law, an e.m.f is induced in a loop of wire if there is a rate of change in flux passing through the wire.

  $inducedemf=-\frac{d\varnothing }{dt}$

Where, $\varnothing$ is the flux which is the dot product of magnetic field and the area A of the loop.

  $\varnothing =\overrightarrow{B}\cdot \overrightarrow{A}$

The induced emf depends on the rate of change in flux. Considering the area of the small loops same, the change in magnetic flux will be same for both the loops. Therefore, the induced emf will be same.

The induced current will be different in both the loops though, as the resistance of the loop 2 is greater than the loop 1.

Conclusion:

Therefore, induced emf will be same in both smaller loops.