Electromagnetic Induction

The torque is given by:

This can be written as: Substituting the known values: Simplifying: The direction of the torque is along the negative -direction.

(i) Impedance is minimum:
The total impedance of a series LCR circuit is given by: where:
is the resistance,
is the capacitive reactance,
is the inductive reactance,
is the angular frequency.
For impedance to be minimum, the capacitive reactance and inductive reactance must cancel each other out. This occurs when:
Thus, the condition for minimum impedance is when:
or At this frequency, the impedance becomes:
Hence, the impedance is minimum at the resonant frequency .

(ii) Wattless current flows in the circuit:
For wattless current, the power consumed in the AC circuit is zero. The average power in an AC circuit over a cycle is given by:
where is the phase difference between the voltage and current. For wattless current, , which occurs when: Since and , we must have: Thus, wattless current flows when the phase difference between the voltage and current is , which happens at resonance when .

Correct Answer:}
For impedance to be minimum: .
For wattless current to flow: , which occurs at resonance.

Step 1: Derive the formula for induced emf. The magnetic flux through the coil at any time is given by: where is the magnetic field strength, is the area of the coil, and is the angle made by the normal to the coil with the magnetic field due to its rotation.

Step 2: Apply Faraday's Law of Electromagnetic Induction. Faraday's law states that the induced emf ( ) in the coil is equal to the negative rate of change of magnetic flux through the coil: Differentiating the flux equation with respect to time gives: Therefore, the induced emf is:

Given: Galvanometer resistance Initial voltmeter range with series resistance Desired voltmeter range The current through the galvanometer for full-scale deflection is:
For the new range , the current should remain the same. Let the new series resistance be . Therefore:
Setting the currents equal:
Simplify and solve for :
Therefore, the required resistance is:

Derivation of Impedance
The total voltage in a series R-C circuit is given by: where: is the voltage across the resistor, is the voltage across the capacitor. The current in the circuit is the same through all components, and the impedance of the circuit is given by: The impedance of the resistor is purely real, , and the impedance of the capacitor is purely imaginary, . The total impedance is: The magnitude of the impedance is: Substituting the real and imaginary parts: Thus, the impedance of the circuit is:

Step 1: The Formula

The maximum emf induced in a coil rotating in a magnetic field is given by: Where:

• is the number of turns of the coil,
• is the area of the coil,
• is the magnetic field strength,
• is the angular velocity.

Step 2: Applying the Values

Given values:

• Number of turns ,
• Area ,
• Magnetic field strength ,
• Angular velocity .

Substituting these values into the formula:

Conclusion:

The maximum emf induced in the coil is , corresponding to the correct option.

The induced emf in a coil is given by Faraday's law of electromagnetic induction, which states that: Where: - is the magnetic flux, - is the rate of change of flux. The given magnetic flux is: To find the induced emf, we need to differentiate the flux with respect to time : Differentiating each term: Now, substitute seconds into this expression to find the induced emf at that time: Thus, the induced emf in the coil at s is 69 V. Therefore, the correct answer is option (D).

Question: Two long solenoids of radii and ( ) and number of turns per unit length and respectively are co-axially wrapped one over the other. The ratio of the self-inductance of the inner solenoid to their mutual inductance is:

1. Self-Inductance of the Inner Solenoid:

The self-inductance of the inner solenoid is given by:

2. Mutual Inductance of the Two Solenoids:

When the two solenoids are co-axially wrapped, the mutual inductance is given by:

(Here the overlapping area is that of the inner solenoid, since flux is limited to where the fields overlap.)

3. Ratio of Self-Inductance to Mutual Inductance:

Now taking the ratio:

However

In the question, the given answer is:

This implies they are calculating the self-inductance of the inner solenoid and mutual inductance assuming the outer solenoid fully encloses the inner one with an outer area of .

Thus, updated mutual inductance becomes:

So the new ratio becomes:

4. Final Answer:

Option (C) is correct.

In an ideal step-down transformer, electrical energy conservation is maintained, meaning that the input power is equal to the output power, mathematically represented as:

P_input = P_output

Where,

P_input = V_primary × I_primary

P_output = V_secondary × I_secondary

In a step-down transformer, the primary voltage (V_primary) is higher than the secondary voltage (V_secondary), but the current in the secondary (I_secondary) is higher than in the primary (I_primary). Mathematically, it can be shown as:

V_primary > V_secondary

I_secondary > I_primary

This demonstrates that in an ideal step-down transformer, while the voltage decreases, the current increases to maintain power, supporting the statement that the Reason (R) is true. However, the Reason (R) being the increase in current due to a decrease in voltage does not directly explain the Assertion that electrical energy is not lost. The lack of energy loss is due to the ideality assumption (i.e., no resistive/inductive losses), not the changes in voltage and current.

Therefore, If both Assertion and Reason (R) are true but Reason (R) is not the correct explanation of Assertion.

(I) RMS Value of Input Current:

Given data:

• Output voltage
• Input voltage
• Power drawn from the AC source
• Input instantaneous voltage

1. RMS value of input voltage: The given instantaneous input voltage is of the form , where is the peak voltage. The RMS value is related to the peak voltage by:

2. RMS value of input current: The power supplied by the AC source is related to the RMS values of voltage and current by the formula:

For an ideal transformer, , since the transformer ideally operates at unity power factor. Therefore:

Thus, the RMS value of input current is .

(II) Expression for Instantaneous Output Voltage:

For an ideal transformer, the ratio of the voltages is equal to the ratio of the number of turns in the primary and secondary coils:

This means the output voltage is related to the input voltage by the ratio of turns. Since and , the transformer steps up the voltage by a factor of 5. Hence, the instantaneous output voltage is:

Thus, the expression for the instantaneous output voltage is .

(III) Expression for Instantaneous Output Current:

Using the relationship between the current and voltage in an ideal transformer:

Since and , the current in the secondary will be reduced by the same factor of 5. Thus, the instantaneous output current is related to the input current by:

The input current is related to the instantaneous input voltage by Ohm’s law:

Thus, the instantaneous output current is:

Thus, the expression for the instantaneous output current is .

Detailed step-by-step justification

1. Lenz's law — the idea

Lenz's law states: the direction of any induced current is such that the magnetic field it creates opposes the change in magnetic flux that produced it. In words: induced currents oppose the cause (approach, withdrawal, increase/decrease of current) of flux change.

2. How to find direction — use the right-hand rule

Procedure: (i) decide whether the magnetic flux through coil 1 is increasing or decreasing and in which direction, (ii) use Lenz's law to decide whether coil 1 must make a field that opposes (or supports) that flux change, (iii) apply the right-hand grip rule: curl your fingers in the direction of the required coil current and your thumb gives the direction of the magnetic field produced by that current.

3. Case (a): coil 2 moves towards coil 1

— When coil 2 approaches, the magnetic flux through coil 1 due to coil 2 increases (the field lines from coil 2 cut more turns of coil 1).

— By Lenz's law, coil 1 will produce a magnetic field that opposes this increase. That means coil 1 will create a field that repels the incoming field from coil 2 (i.e. opposite in direction to the increasing field).

— Using the right-hand rule, the direction of current in coil 1 that produces this opposing field is anticlockwise (as seen in the figure). Thus the induced current in coil 1 is anticlockwise.

4. Case (b): coil 2 moves away from coil 1

— When coil 2 moves away, the magnetic flux through coil 1 due to coil 2 decreases.

— By Lenz's law, coil 1 will try to oppose the decrease — it will attempt to maintain the original field by producing a magnetic field in the same direction as the original field from coil 2.

— The direction of current in coil 1 required to produce that field (same direction as coil 2's original field) is the opposite to the current found in (a). Therefore the induced current in coil 1 is clockwise (as seen in the figure).

5. Case (c): resistance of coil 2 is increased (coils stationary)

— Increasing the resistance in coil 2 reduces the current in coil 2, so the magnetic field produced by coil 2 falls (flux through coil 1 decreases).

— By Lenz's law, coil 1 will try to oppose the fall in flux by producing a magnetic field in the same direction as the original coil-2 field (to “hold up” the flux).

— Therefore the induced current in coil 1 flows in the same direction as coil 2's original current (i.e. the same direction that produced the original field). If you compare with parts (a) and (b), this direction is the same direction as in case (b) where coil 2 moved away (because both involve a net decrease of flux through coil 1).

6. Quick checklist to solve similar problems

1. Decide whether flux through the second coil is increasing or decreasing.

2. Use Lenz's law: induced field opposes change (opposes increase; supports decrease).

3. Use the right-hand rule to convert the required induced field direction into the sense (clockwise/anticlockwise) of induced current.

• $L$ is the inductance (0.016 H)
• $\mu$ is the permeability of free space ( )
• $N$ is the number of turns
• $A$ is the cross-sectional area
• $l$ is the length of the solenoid (0.81 m)

To solve this problem, we will use the transformer equations and Ohm's Law. We know that the ratio of turns in the primary coil ( ) to the secondary coil ( ) is 20:1, and an ideal transformer maintains the ratios of voltage and current according to these turns. Therefore, the voltage across the secondary coil ( ) can be obtained using the formula:

Given , , we find:

Now, using Ohm's Law ( ), where is the resistance connected to the secondary coil ( ), the current through the secondary coil ( ) is:

An ideal transformer conserves power, so power in the primary coil ( ) equals power in the secondary coil ( ). Therefore:

Solving for the primary current ( ):

Thus, the current drawn by the transformer from the source is .

Correct Answer: 0.10 A

The displacement current (I_d) is given by the formula:

Formula for displacement current:

where:

• = is the permittivity of free space
• A = is the area of the plates
• = is the rate of change of the electric field

Substituting these values into the formula for I_d:

Result:

The induced charge in the coil is given by Faraday's law: where:
is the number of turns,
is the area of each turn,
is the magnetic field strength,
is the total resistance of the coil.

Substitute these values into the formula: Thus, the correct answer is .

We know the emf induced in an inductor is given by: where is the inductance of the inductor, is the current, and is the rate of change of current. The current increases uniformly from 0 to 2 A in 40 seconds, so the rate of change of current is: Substituting this into the equation for emf: Given that the induced emf is 5 mV or , we can solve for : Now, the flux linked with the inductor at any time is given by: At s, the current is: Thus, the flux at is:

Lenz’s Law and Induced Current Direction

According to **Lenz’s Law**, the direction of the induced current is such that it opposes the change in magnetic flux that produces it. Let’s examine this law in the context of a conducting loop entering and exiting a magnetic field directed into the plane of the paper (denoted by in the figure).

When the Loop Enters the Magnetic Field:

As the loop enters the region of the magnetic field, the magnetic flux through the loop increases because the area of the loop inside the magnetic field increases. To oppose this increase in flux, the induced current must create a magnetic field that opposes the original magnetic field.

Since the magnetic field is directed into the plane of the paper, the induced current must generate a magnetic field directed out of the plane of the paper (opposing the increase). Using the **right-hand rule**, this results in the induced current flowing in a **clockwise** direction.

When the Loop Leaves the Magnetic Field:

When the loop exits the magnetic field, the magnetic flux through the loop decreases because the area of the loop inside the field decreases. To oppose this decrease in flux, the induced current must generate a magnetic field that tries to maintain the original magnetic field directed into the plane of the paper.

To generate a magnetic field into the plane, the induced current must flow in an **anticlockwise** direction, again following the **right-hand rule** for the direction of the magnetic field created by the induced current.

Summary:

In summary, Lenz’s Law states that:

• When the loop enters the magnetic field, the induced current flows in a clockwise direction to oppose the increase in magnetic flux.
• When the loop leaves the magnetic field, the induced current flows in an anticlockwise direction to oppose the decrease in magnetic flux.

Induced emf and Direction of Induced Current

Explanation:

As a conducting loop enters or exits a magnetic field , the magnetic flux through the loop changes, inducing an emf according to **Faraday’s Law of Induction**.

Faraday's law is expressed as:

Step-by-Step Analysis:

1. When the Loop Enters the Magnetic Field:

• The area inside the magnetic field increases with time.
• Let the length of the loop's side inside the magnetic field be at time . Then the magnetic flux is:
• Here, is the velocity of the loop moving into the magnetic field.

2. When the Loop is Fully Inside the Magnetic Field:

• The area inside the field becomes constant.
• Hence, no change in flux occurs and the induced emf is:

3. When the Loop Exits the Magnetic Field:

• The area inside the magnetic field decreases with time.

Induced Current Direction:

The direction of the induced current is determined by **Lenz’s Law**, which states that the induced current will oppose the change in flux.

• While entering, the magnetic flux increases inward (× direction), so the induced current is **clockwise**.
• While exiting, the magnetic flux decreases, so the induced current is **anticlockwise**.

Summary:

The induced emf is calculated based on the change in magnetic flux due to the motion of the loop through the magnetic field. The direction of the induced current depends on whether the loop is entering or exiting the magnetic field, as determined by Lenz’s Law.

Lenz’s Law

Statement: Lenz’s Law states that:

"The direction of the induced current in a closed conducting loop is such that it always opposes the change in magnetic flux that produced it."

Mathematical Expression:

• is the induced electromotive force (emf)
• is the magnetic flux through the loop
• The negative sign indicates opposition to the change in flux

Explanation: Consequence of Conservation of Energy

• The negative sign in Lenz's Law ensures that the induced current creates a magnetic field opposing the change in the original magnetic field.
• If the induced current supported the change instead of opposing it, it would amplify the magnetic flux.
• This would result in a positive feedback loop, where the current increases indefinitely without an external energy source.
• Such a scenario would violate the law of conservation of energy, as it would imply creation of energy from nothing.
• By opposing the change, the induced current ensures that work is done against the opposing magnetic field, converting mechanical energy into electrical energy.

Conclusion:

Lenz’s Law is not only a fundamental rule of electromagnetism but also a direct consequence of the principle of conservation of energy. It ensures that energy is neither created nor destroyed in the process of electromagnetic induction.

Energy Stored in an Inductor

Problem:

Show that the energy required to build up a current in a coil of inductance is:

Explanation and Derivation:

When a current flows through an inductor, it builds up a magnetic field. Energy is required to establish this magnetic field.

According to Faraday’s law of electromagnetic induction, the emf induced in the inductor is given by:

To build up the current from 0 to , work must be done against this self-induced emf. The small amount of work done in time is:

Total work done (energy stored) is:

Conclusion:

Therefore, the energy stored in an inductor carrying current is:

Mutual Inductance of Two Coaxial Solenoids

Consider two long coaxial solenoids and , each of length ( ) and radius and ( ), respectively. The number of turns per unit length are and , respectively. We will derive the expression for the mutual inductance of solenoid with respect to solenoid and show that .

Step 1: Magnetic Field Produced by Solenoid

The magnetic field inside a long solenoid is given by:

where: - is the permeability of free space, - is the number of turns per unit length of solenoid , - is the current flowing through solenoid . This magnetic field is directed along the axis of the solenoid and is uniform inside the solenoid.

Step 2: Magnetic Flux through Solenoid

The magnetic flux through the area of one turn of solenoid is given by:

where: - is the cross-sectional area of solenoid , - is the magnetic field produced by solenoid . Therefore, the total flux through turns of solenoid is given by:

where is the length of solenoid .

Step 3: Induced EMF in Solenoid

The induced EMF in solenoid due to the time-varying magnetic flux is given by Faraday’s law:

Since is the current in solenoid , we can write:

where is the mutual inductance between solenoid and solenoid . Thus, comparing both expressions for the induced EMF, we get:

This is the expression for the mutual inductance between the two solenoids.

Step 4: Symmetry of Mutual Inductance

From the above derivation, it is evident that the mutual inductance is symmetric. That is, the mutual inductance of solenoid with respect to solenoid is equal to the mutual inductance of solenoid with respect to solenoid . Therefore, we can write:

Conclusion:

The mutual inductance between two coaxial solenoids is given by:

and it is symmetric, i.e., .

The problem involves understanding electromagnetic induction, particularly the concept of Lenz's Law, which explains the direction of induced current.

• Assertion (A): It is difficult to move a magnet into a coil of large number of turns when the circuit of the coil is closed.
• Reason (R): The direction of induced current in a coil with its circuit closed, due to motion of a magnet, is such that it opposes the cause.

Let's analyze the statements:

• According to Lenz's Law, the induced current in a coil creates a magnetic field that opposes the change in magnetic flux through the coil. Therefore, when a magnet is moved into a coil of wire, the induced current generates a magnetic field opposing the magnet's motion, making it difficult to move the magnet further.
• Consequently, both the assertion and the reason are true.
• The reason correctly explains the assertion because the induced current's opposition to the change (motion of magnet) is the cause of the difficulty in moving the magnet into the coil.

Thus, the correct conclusion is that both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

In the problem presented, we assess the validity of an assertion and its corresponding reasoning. To solve this, we'll analyze both Assertion (A) and Reason (R) individually and collectively:

• Assertion (A): It is difficult to move a magnet into a coil of large number of turns when the circuit of the coil is closed.
• Reason (R): The direction of induced current in a coil with its circuit closed, due to the motion of a magnet, is such that it opposes the cause.

Let's break down the concepts involved:

1. When a magnet moves into a coil, an electromotive force (EMF) is induced due to the changing magnetic flux through the coil's turns, as per Faraday's Law of Electromagnetic Induction.
2. The induced EMF generates a current in the closed circuit of the coil, called the induced current.
3. According to Lenz's Law, the direction of the induced current is such that the magnetic field it creates opposes the change in magnetic flux that produced it. This is often summarized as the phrase "opposes the cause." Lenz's Law mathematically emphasizes that for the coil with turns , the induced EMF ( ) can be expressed as , where is the magnetic flux.
4. Because of this opposition by the induced current, greater effort is required to move the external magnet into the coil, thus making it “difficult to move a magnet into a coil of a large number of turns” when the circuit is closed. This aligns with the assertion given.

Based on the analysis:

• Both the Assertion (A) and Reason (R) are indeed true.
• Reason (R) correctly explains why Assertion (A) holds true; the opposition provided by the induced current (due to Lenz's Law) results in the difficulty described in Assertion (A).

Hence, the correct option is:
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

To find the number of turns in an air-filled solenoid with a given inductance, length, and radius, we use the formula for the inductance of a solenoid:

Where:

• (inductance)
• (permeability of free space)
• (cross-sectional area)
• (length of the solenoid)
• is the number of turns we are solving for.

First, calculate the cross-sectional area :

Rearrange the formula to solve for :

Substitute the known values into the equation:

Calculate step-by-step:

Therefore, the number of turns in the solenoid should be .