Motional EMF in Rotating and Sliding Systems under Gravity: Calculating the terminal velocity, induced current, and rate of heat dissipation for a conducting rod sliding down parallel vertical rails under gravity in a uniform magnetic field.

1. Introduction

When a conducting rod moves through a magnetic field, the free charges inside the rod experience a magnetic force. This separation of charges produces an induced emf called motional EMF.

Basic idea:
Moving conductor + Magnetic field → Motional EMF → Current (if circuit is closed)

A very important application is a conducting rod sliding on two parallel conducting rails under gravity in a uniform magnetic field.

2. Motional EMF

For a rod of length L moving with velocity v perpendicular to a uniform magnetic field B:

ε = BLv

Where:

• B = magnetic field
• L = length of conducting rod
• v = velocity of rod
• ε = induced EMF

The faster the rod moves, the larger the induced EMF.

3. Why is EMF Produced?

A charge q moving with velocity v in magnetic field B experiences the magnetic Lorentz force:

F = qvB

This force separates positive and negative charges along the rod. Consequently, a potential difference develops between its ends.

ε = BLv

4. Conducting Rod Sliding Down Rails

Consider a conducting rod of mass m and length L placed on two parallel vertical rails. The rod is free to slide downward under gravity.

A uniform magnetic field B is perpendicular to the plane of the rails. The circuit has total resistance R.

The important quantities are:

• Weight = mg
• Motional EMF = BLv
• Induced current = BLv/R
• Magnetic force = BIL

5. Induced Current

Using Ohm's law:

I = ε/R

Since:

ε = BLv

Therefore:

I = BLv/R

Thus induced current increases with velocity.

6. Magnetic Force on the Rod

A current-carrying conductor in a magnetic field experiences force:

F = BIL

Substituting:

I = BLv/R

We get:

F_B = B²L²v/R

According to Lenz's law, this magnetic force opposes the downward motion.

Magnetic force acts upward when the rod is sliding downward.

7. Equation of Motion

The downward gravitational force is:

mg

The upward magnetic force is:

B²L²v/R

Therefore:

m dv/dt = mg - B²L²v/R

or:

m dv/dt = mg - (B²L²/R)v

8. Terminal Velocity

At terminal velocity, acceleration becomes zero:

dv/dt = 0

Therefore:

mg = B²L²v_t/R

Solving:

v_t = mgR/(B²L²)

Important:
Terminal velocity increases with mass m and resistance R, but decreases with B² and L².

9. Current at Terminal Velocity

The current is:

I = BLv/R

Putting:

v = mgR/(B²L²)

we get:

I_t = BL/R × mgR/(B²L²)

Therefore:

I_t = mg/(BL)

This result is important for competitive examinations.

10. Rate of Heat Dissipation

Electrical power converted into heat in the circuit is:

P = I²R

Since:

I = BLv/R

Therefore:

P = B²L²v²/R

At terminal velocity:

P_t = m²g²R/(B²L²)

11. Energy Conservation

When the rod falls through a vertical distance, gravitational potential energy is converted into:

• Kinetic energy
• Electrical energy
• Heat energy

At terminal velocity, kinetic energy does not increase because velocity remains constant.

Therefore:

mgv_t = I_t²R

This confirms that gravitational power is completely converted into electrical heating at terminal velocity.

12. Power Balance at Terminal Velocity

Mechanical power supplied by gravity:

P_gravity = mgv_t

Electrical heating:

P_heat = I_t²R

Since:

I_t=mg/(BL)

and:

v_t=mgR/(B²L²)

Both expressions give:

P_gravity = P_heat

13. Velocity as a Function of Time

The equation is:

m dv/dt = mg - (B²L²/R)v

Define:

k = B²L²/(mR)

Then:

dv/dt = g-kv

If the rod starts from rest:

v(t) = v_t(1-e^-kt)

where:

v_t=mgR/(B²L²)

and:

k = B²L²/(mR)

14. Time Constant

The time constant of the motion is:

τ = mR/(B²L²)

The velocity can be written as:

v = v_t(1-e^-t/τ)

After a few time constants, the rod approaches terminal velocity.

15. Rotating Conducting Rod

For a conducting rod of length L rotating with angular velocity ω in a uniform magnetic field perpendicular to the plane of rotation, different parts of the rod have different linear velocities.

At a distance r from the axis:

v = ωr

The small EMF generated in length dr is:

dε = Bv dr

Therefore:

dε = Bωr dr

Integrating from r=0 to r=L:

ε = ∫₀ᴸ Bωr dr

Hence:

ε = ½BωL²

16. Sliding Rod vs Rotating Rod

System	Velocity	Motional EMF
Sliding rod	v	BLv
Rotating rod	ωr	½BωL²

17. Solved Example — Terminal Velocity

A conducting rod of mass 0.5 kg and length 0.5 m slides vertically on conducting rails. The magnetic field is 1 T and total resistance is 2 Ω. Find terminal velocity. Take g=10 m/s².

v_t=mgR/(B²L²)

v_t = (0.5)(10)(2) / (1²)(0.5²)

v_t=40 m/s

18. Solved Example — Terminal Current

For a rod of mass 2 kg, length 1 m in a magnetic field of 0.5 T, find the terminal current. Take g=10 m/s².

I_t=mg/(BL)

I_t = (2)(10)/(0.5)(1)

I_t=40 A

19. Solved Example — Heat Dissipation

A rod moves with velocity 5 m/s in a magnetic field of 2 T. Its length is 0.4 m and circuit resistance is 4 Ω. Find the rate of heat production.

P=B²L²v²/R

P = (2²)(0.4²)(5²)/4

P=4 W

20. Solved Example — Rotating Rod

A conducting rod of length 0.4 m rotates with angular speed 100 rad/s in a magnetic field of 0.5 T. Find the induced EMF.

ε=½BωL²

ε = ½(0.5)(100)(0.4²)

ε=4 V

21. Important Formula Sheet

Quantity	Formula
Motional EMF	ε=BLv
Induced current	I=BLv/R
Magnetic force	F=BIL
Magnetic force in terms of v	F=B²L²v/R
Equation of motion	m dv/dt=mg−B²L²v/R
Terminal velocity	vt=mgR/(B²L²)
Terminal current	It=mg/(BL)
Heat dissipation	P=I²R=B²L²v²/R
Terminal heat power	Pt=m²g²R/(B²L²)
Time constant	τ=mR/(B²L²)
Rotating rod EMF	ε=½BωL²

Section A — 1 Mark MCQs
10 Questions

Q1. The motional EMF in a rod of length L moving with speed v perpendicular to B is: 1 Mark

A) B/Lv

B) BLv

C) Bv/L

D) Lv/B

Answer: B

Q2. The induced current in a circuit of resistance R is: 1 Mark

A) BLvR

B) BLv/R

C) R/BLv

D) B/RLv

Answer: B

Q3. The magnetic force on the sliding rod is: 1 Mark

A) BIL

B) BI/L

C) BL/I

D) B/I L

Answer: A

Q4. At terminal velocity, acceleration is: 1 Mark

A) g

B) 2g

C) Zero

D) Infinite

Answer: C

Q5. Terminal velocity of the rod is proportional to: 1 Mark

A) B²

B) 1/R

C) mR

D) 1/m

Answer: C

Q6. The magnetic force on a falling rod acts: 1 Mark

A) Downward

B) Upward

C) Horizontally only

D) Zero always

Answer: B

Q7. At terminal velocity, gravitational force is equal to: 1 Mark

A) Electric force

B) Magnetic force

C) Normal force only

D) Zero

Answer: B

Q8. The rotating rod of length L has motional EMF: 1 Mark

A) BωL

B) BωL²

C) ½BωL²

D) 2BωL²

Answer: C

Q9. Electrical power dissipated as heat is: 1 Mark

A) IR

B) I²R

C) I/R

D) R/I

Answer: B

Q10. At terminal velocity, gravitational power is converted into: 1 Mark

A) Only kinetic energy

B) Electrical heating

C) Potential energy

D) Nuclear energy

Answer: B

Section B — 2 Marks
10 Questions

Q11. Define motional EMF. 2 Marks

The EMF induced in a conductor due to its motion through a magnetic field is called motional EMF. For perpendicular motion:

ε=BLv

Q12. Write the expression for induced current in the sliding rod. 2 Marks

I=BLv/R

Q13. Why does the magnetic force oppose the motion of the rod? 2 Marks

According to Lenz's law, the induced current produces a magnetic effect that opposes the change responsible for producing it. Hence the magnetic force opposes the motion of the rod.

Q14. What is terminal velocity? 2 Marks

Terminal velocity is the constant velocity attained when the downward gravitational force becomes equal to the upward magnetic force.

Q15. Write the terminal velocity of the rod. 2 Marks

v_t=mgR/(B²L²)

Q16. Write the terminal current. 2 Marks

I_t=mg/(BL)

Q17. Write the expression for heat dissipation in the circuit. 2 Marks

P=I²R

Q18. How does terminal velocity change if magnetic field is doubled? 2 Marks

Since:

v_t∝1/B²

If B is doubled, terminal velocity becomes one-fourth.

Q19. How does terminal velocity change if the resistance is doubled? 2 Marks

Since:

v_t∝R

Doubling R doubles the terminal velocity.

Q20. Why does the rod not accelerate indefinitely? 2 Marks

As velocity increases, induced current and magnetic opposing force increase. Eventually magnetic force becomes equal to weight, making acceleration zero.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the terminal velocity of a conducting rod sliding under gravity. 3 Marks

Motional EMF:

ε=BLv

Current:

I=BLv/R

Magnetic force:

F=BIL=B²L²v/R

At terminal velocity:

mg=B²L²v_t/R

Therefore:

v_t=mgR/(B²L²)

Q22. Derive the terminal current in the sliding rod. 3 Marks

We know:

I=BLv/R

At terminal velocity:

v_t=mgR/(B²L²)

Therefore:

I_t = BL/R × mgR/(B²L²)

Hence:

I_t=mg/(BL)

Q23. Derive the rate of heat dissipation in terms of velocity. 3 Marks

Electrical power:

P=I²R

Current:

I=BLv/R

Therefore:

P=(BLv/R)²R

Hence:

P=B²L²v²/R

Q24. Show that at terminal velocity, mechanical power supplied by gravity equals electrical heat power. 3 Marks

Mechanical power:

P_g=mgv_t

At terminal velocity:

mg=BIL

Multiplying both sides by v:

mgv=BILv

Since:

BLv=IR

Therefore:

mgv=I²R

Hence:

P_gravity=P_heat

Q25. A rod of mass 1 kg and length 0.5 m moves on rails in a magnetic field of 2 T. Total resistance is 4 Ω. Find terminal velocity. Take g=10 m/s². 3 Marks

v_t=mgR/(B²L²)

v_t = (1)(10)(4) / (2²)(0.5²)

v_t=20 m/s

Q26. A rod of mass 2 kg and length 0.5 m moves in a magnetic field of 1 T. Find terminal current. Take g=10 m/s². 3 Marks

I_t=mg/(BL)

I_t = (2)(10)/(1)(0.5)

I_t=40 A

Q27. A rod of length 0.5 m moves at 10 m/s in a 2 T magnetic field. If total resistance is 5 Ω, calculate induced EMF and current. 3 Marks

EMF:

ε=BLv

ε=(2)(0.5)(10)=10 V

Current:

I=ε/R=10/5

ε=10 V,   I=2 A

Q28. A rotating conducting rod of length 0.6 m rotates with angular speed 50 rad/s in a magnetic field of 0.8 T. Find the induced EMF. 3 Marks

ε=½BωL²

ε = ½(0.8)(50)(0.6²)

ε=7.2 V

Q29. A rod of mass 0.5 kg, length 1 m and resistance 2 Ω moves vertically in a magnetic field of 1 T. Find terminal velocity and terminal current. Take g=10 m/s². 3 Marks

Terminal velocity:

v_{t}=mgR/(B²L²)}

v_t = (0.5)(10)(2)/(1²)(1²) =10 m/s

Terminal current:

I_t=mg/(BL)

I_t=(0.5)(10)/(1)(1) =5 A

v_t=10 m/s,   I_t=5 A

Q30. Explain why increasing the magnetic field decreases terminal velocity but increases the magnetic opposing force. 3 Marks

Terminal velocity:

v_t∝1/B²

Therefore increasing B decreases terminal velocity. But magnetic force is:

F_B=B²L²v/R

Hence, for a given velocity, increasing B increases the magnetic force. At terminal velocity this increased magnetic force balances the same weight mg.

22. Competitive Exam Quick Tricks

1. Sliding rod:

ε=BLv

2. Current:

I=BLv/R

3. Magnetic force:

F=B²L²v/R

4. Terminal velocity:

v_t=mgR/(B²L²)

5. Terminal current:

I_t=mg/(BL)

6. Heat power:

P=I²R=B²L²v²/R

7. At terminal velocity:

mgv_t=I_t²R

8. Rotating rod:

ε=½BωL²

Motional EMF in Rotating and Sliding Systems under Gravity
CBSE + JEE/NEET Competitive Physics | Theory + Derivation + Numerical Practice

Electromagnetic Damping in Galvanometers: Mathematical modeling of the electromagnetic damping/restoring torque and its impact on the current and voltage sensitivity of a moving coil galvanometer.

1. Introduction

A moving coil galvanometer is used to detect and measure small electric currents. When current flows through its coil in a magnetic field, the coil experiences a magnetic torque and starts rotating.

During motion, electromagnetic effects can produce a damping torque that opposes the motion of the coil.

Key idea: Damping reduces oscillations of the coil and helps the pointer reach its final position quickly.

2. Principle of Moving Coil Galvanometer

For a coil of N turns, area A, carrying current I in a radial magnetic field B, the deflecting torque is:

τ_d = NBIA

The restoring torque of the spring is:

τ_r = Cθ

where C is the torsional constant and θ is angular deflection.

At equilibrium:

NBIA = Cθ

Therefore:

θ = (NBA/C) I

Hence:

I/θ = C/(NBA)

3. What is Electromagnetic Damping?

When the coil moves in the magnetic field, the magnetic flux linked with the coil changes. According to Faraday's law, an induced emf is produced.

e = -N dΦ/dt

The induced current produces a magnetic torque which opposes the motion according to Lenz's law.

τ_damp = -K_dω

where:

• K_d = electromagnetic damping constant
• ω = angular velocity

The negative sign shows that the damping torque is opposite to the direction of angular motion.

4. Equation of Motion of the Coil

Let:

• J = moment of inertia of the coil
• C = torsional constant
• K_d = damping constant
• θ = angular displacement

The equation of rotational motion is:

J(d²θ/dt²) + K_d(dθ/dt) + Cθ = NBIA

This is mathematically similar to a damped harmonic oscillator.

5. Free Motion After Current is Removed

If the current is suddenly removed:

NBIA = 0

The equation becomes:

Jθ̈ + K_dθ̇ + Cθ = 0

The coil can undergo damped oscillations before settling at its final position.

6. Damping Ratio

For the equation:

Jθ̈ + K_dθ̇ + Cθ = 0

The natural angular frequency is:

ω₀ = √(C/J)

The damping ratio is:

ζ = K_d / (2√(JC))

7. Types of Damping

Condition	Type	Behaviour
ζ < 1	Underdamped	Oscillations occur
ζ = 1	Critically damped	Fastest settling without oscillation
ζ > 1	Overdamped	Slow settling without oscillation

For a galvanometer, suitable damping allows the pointer to settle quickly without excessive oscillation.

8. Mathematical Model of Electromagnetic Damping

Suppose the coil rotates with angular velocity ω. The rate of change of magnetic flux produces an induced emf:

e ∝ ω

If the electrical resistance of the circuit is R:

i_ind = e/R

Therefore:

i_ind ∝ ω

The magnetic torque due to this current is:

τ_damp ∝ i_ind

Hence:

τ_damp = -K_dω

Thus electromagnetic damping torque is proportional to angular velocity.

9. Current Sensitivity

Current sensitivity is the angular deflection produced per unit current.

S_I = θ/I

For a moving coil galvanometer:

θ/I = NBA/C

Therefore:

S_I = NBA/C

Current sensitivity increases when:

• Number of turns N increases
• Area A increases
• Magnetic field B increases
• Torsional constant C decreases

10. Voltage Sensitivity

Voltage sensitivity is angular deflection per unit applied voltage.

S_V = θ/V

If the galvanometer resistance is G:

V = IG

Therefore:

θ/V = (θ/I)/G

Hence:

S_V = NBA/(CG)

11. Effect of Damping on Sensitivity

For an ideal steady-state measurement:

θ = NBAI/C

Therefore the static current sensitivity is:

S_I=NBA/C

Electromagnetic damping mainly affects the transient response and settling time. It does not directly alter the static deflection relation when the coil has reached equilibrium.

Exam point: Do not incorrectly conclude that increasing damping automatically reduces the final static sensitivity. Damping primarily controls oscillation and settling behaviour.

12. Effect of Circuit Resistance on Electromagnetic Damping

Since:

i_ind=e/R

a larger resistance reduces the induced current. Therefore the electromagnetic damping becomes weaker.

Approximately:

K_d ∝ 1/R

for a simple resistive damping model.

Lower resistance → larger induced current → stronger electromagnetic damping.

13. Short-Circuit Damping

If the galvanometer circuit is effectively short-circuited, the resistance of the path is small.

Therefore the induced current is large:

i_ind=e/R

Hence:

τ_damp ∝ i_ind

Strong electromagnetic damping occurs.

This is why short-circuiting a moving coil galvanometer can make its pointer settle rapidly.

14. Settling Behaviour

The damping determines how quickly the pointer reaches its final equilibrium position.

Jθ̈ + K_dθ̇ + Cθ = NBIA

For stronger appropriate damping:

• Oscillations decrease
• Overshoot decreases
• Pointer settles faster

Critical damping gives the fastest non-oscillatory response in the ideal second-order model.

15. Solved Example 1 — Current Sensitivity

A galvanometer has N = 100 turns, area A = 2×10^-4 m², magnetic field B = 0.5 T and torsional constant C = 10^-6 N m rad^-1. Find current sensitivity.

S_I=NBA/C

S_I = (100)(0.5)(2×10^-4) / 10^-6

S_I=10⁴ rad/A

16. Solved Example 2 — Voltage Sensitivity

If the current sensitivity of a galvanometer is 5×10³ rad/A and its resistance is 50 Ω, find its voltage sensitivity.

S_V=S_I/G

S_V = (5×10³)/50

S_V=100 rad/V

17. Solved Example 3 — Damping Torque

A galvanometer has damping constant K_d=2×10^-6 N m s rad^-1. If its angular velocity is 50 rad/s, find the damping torque.

τ_damp=-K_dω

τ_damp = -(2×10^-6)(50)

τ_damp=-1×10^-4 N m

The negative sign indicates opposition to motion.

18. Solved Example 4 — Critical Damping

For a galvanometer:

• J = 4×10^-6 kg m²
• C = 10^-4 N m rad^-1

Find the critical damping constant.

For critical damping:

K_d,critical=2√(JC)

Therefore:

K_d,critical = 2√[(4×10^-6)(10^-4)]

K_d,critical=4×10^-5 N m s rad^-1

19. Important Formula Sheet

Quantity	Formula
Deflecting torque	τd=NBIA
Restoring torque	τr=Cθ
Equilibrium	NBIA=Cθ
Current sensitivity	SI=θ/I=NBA/C
Voltage sensitivity	SV=θ/V=NBA/(CG)
Damping torque	τdamp=-Kdω
Equation of motion	Jθ̈+Kdθ̇+Cθ=NBIA
Natural frequency	ω₀=√(C/J)
Damping ratio	ζ=Kd/(2√JC)
Critical damping	Kd=2√JC

Section A — 1 Mark MCQs
10 Questions

Q1. The restoring torque in a moving coil galvanometer is: 1 Mark

A) NBIA

B) Cθ

C) IR

D) BIL

Answer: B

Q2. Electromagnetic damping torque is proportional to: 1 Mark

A) θ

B) I

C) Angular velocity

D) Resistance only

Answer: C

Q3. Current sensitivity of a galvanometer is: 1 Mark

A) I/θ

B) θ/I

C) V/θ

D) θV

Answer: B

Q4. Current sensitivity is equal to: 1 Mark

A) C/NBA

B) NBA/C

C) NBA/C²

D) C/NB

Answer: B

Q5. Voltage sensitivity is: 1 Mark

A) S_IG

B) S_I/G

C) G/S_I

D) S_I+G

Answer: B

Q6. In electromagnetic damping, the induced current opposes: 1 Mark

A) Applied voltage

B) Coil motion

C) Spring force

D) Resistance

Answer: B

Q7. Critical damping corresponds to: 1 Mark

A) ζ=0

B) ζ<1

C) ζ=1

D) ζ>1 only

Answer: C

Q8. The damping torque is written as: 1 Mark

A) K_dω

B) -K_dω

C) K_dθ

D) -Cθ

Answer: B

Q9. Increasing the number of turns N generally increases: 1 Mark

A) Current sensitivity

B) Resistance only

C) Damping ratio only

D) Mass only

Answer: A

Q10. Strong electromagnetic damping requires the induced current to be: 1 Mark

A) Small

B) Large

C) Zero

D) Constant zero

Answer: B

Section B — 2 Marks
10 Questions

Q11. Define electromagnetic damping. 2 Marks

Electromagnetic damping is the opposing torque produced by an induced current when the coil of a galvanometer moves in a magnetic field.

Q12. Write the expression for damping torque. 2 Marks

τ_damp=-K_dω

Q13. Write the equation of motion of a damped galvanometer. 2 Marks

Jθ̈+K_dθ̇+Cθ=NBIA

Q14. Define current sensitivity. 2 Marks

Current sensitivity is the angular deflection produced per unit current:

S_I=θ/I

Q15. Define voltage sensitivity. 2 Marks

Voltage sensitivity is angular deflection produced per unit voltage:

S_V=θ/V

Q16. Why is a galvanometer damped? 2 Marks

To reduce oscillations and make the pointer reach its final position quickly and smoothly.

Q17. What happens to electromagnetic damping when circuit resistance decreases? 2 Marks

Induced current increases, so electromagnetic damping becomes stronger.

Q18. Write the relation between current and voltage sensitivity. 2 Marks

If G is galvanometer resistance:

S_V=S_I/G

Q19. What is critical damping? 2 Marks

Critical damping is the condition in which the pointer returns to equilibrium in the shortest time without oscillating.

Q20. Does damping directly determine the final static deflection? 2 Marks

No. For steady current, damping torque becomes zero because angular velocity becomes zero. The final deflection is determined by the balance between magnetic and restoring torques.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the current sensitivity of a moving coil galvanometer. 3 Marks

Deflecting torque:

τ_d=NBIA

Restoring torque:

τ_r=Cθ

At equilibrium:

NBIA=Cθ

Therefore:

θ/I=NBA/C

Hence: Current sensitivity = NBA/C.

Q22. Derive the voltage sensitivity of a galvanometer. 3 Marks

Current sensitivity:

S_I=θ/I=NBA/C

Since:

V=IG

Therefore:

θ/V=(θ/I)/G

Hence:

S_V=NBA/(CG)

Q23. Explain mathematically how electromagnetic damping torque is proportional to angular velocity. 3 Marks

When the coil rotates:

e∝dθ/dt

Since:

ω=dθ/dt

we get:

e∝ω

Induced current:

i=e/R∝ω

Magnetic torque:

τ∝i

Therefore:

τ_damp=-K_dω

Q24. Write and explain the differential equation of a damped galvanometer. 3 Marks

The three torques are:

• Inertial torque = Jθ̈
• Damping torque = K_dθ̇
• Restoring torque = Cθ

Magnetic driving torque:

NBIA

Thus:

Jθ̈+K_dθ̇+Cθ=NBIA

This equation describes the transient and steady-state motion of the pointer.

Q25. A galvanometer has N=200, A=5×10⁻⁴ m², B=0.4 T and C=2×10⁻⁶ N m/rad. Find its current sensitivity. 3 Marks

S_I=NBA/C

S_I = (200)(0.4)(5×10^{-4}) / (2×10-6)}

S_I=2×10⁴ rad/A

Q26. A galvanometer has current sensitivity 2000 rad/A and resistance 100 Ω. Find its voltage sensitivity. 3 Marks

S_V=S_I/G

S_V=2000/100

S_V=20 rad/V

Q27. Explain why short-circuiting a galvanometer produces strong damping. 3 Marks

When the coil moves, an induced emf is generated:

e∝ω

Short circuiting reduces the effective resistance R. Therefore:

i=e/R

The induced current becomes large. Hence:

τ_damp∝i

Strong damping is produced and the pointer settles rapidly.

Q28. For a galvanometer J=10⁻⁵ kg m² and C=4×10⁻⁴ N m/rad. Calculate the critical damping constant. 3 Marks

Critical damping:

K_d,c=2√(JC)

=2√[(10^-5)(4×10^-4)]

K_d,c=4×10^-4 N m s rad^-1

Q29. A galvanometer has sensitivity θ/I = 5000 rad/A. Its resistance is 200 Ω. Find voltage sensitivity. 3 Marks

S_V=S_I/G

S_V=5000/200

S_V=25 rad/V

Q30. Explain the effect of damping on the response of a galvanometer. Also distinguish underdamped, critically damped and overdamped motion. 3 Marks

The equation is:

Jθ̈+K_dθ̇+Cθ=0

The damping ratio is:

ζ=K_d/(2√JC)

• ζ<1: Underdamped — oscillations occur.
• ζ=1: Critically damped — fastest response without oscillation.
• ζ>1: Overdamped — no oscillation but response is slower.

Suitable damping makes the galvanometer fast, stable and easy to read.

Final Quick Revision

τ_d=NBIA

τ_r=Cθ

τ_damp=-K_dω

Jθ̈+K_dθ̇+Cθ=NBIA

S_I=NBA/C

S_V=NBA/(CG)

ω₀=√(C/J)

ζ=K_d/(2√JC)

K_d,critical=2√JC

Competitive Exam Trick:

1. Static sensitivity: use θ/I = NBA/C.
2. Voltage sensitivity: divide current sensitivity by galvanometer resistance: S_V=S_I/G.
3. Damping torque: τ_damp=−K_dω.
4. Transient motion: Jθ̈+K_dθ̇+Cθ=NBIA.
5. Critical damping: K_d=2√JC.

Electromagnetic Damping in Galvanometers
CBSE + Competitive Physics | Theory + Derivation + Solved Examples + 30 Questions

Magnetic Fields of Intersecting and Coaxial Loops: Vector analysis of the net magnetic field magnitude and direction at the center or on the axis of two mutually perpendicular or coaxial current-carrying loops

1. Introduction

A current-carrying circular loop produces a magnetic field. The direction of this field is determined using the right-hand thumb rule.

B = μ₀I / 2R

This is the magnetic field at the centre of a circular loop of radius R carrying current I.

When two loops are present, their magnetic fields must be treated as vectors.

Important: Magnetic fields do not simply add as ordinary numbers when the fields have different directions. First resolve the fields into components and then add them vectorially.

2. Magnetic Field at the Centre of a Circular Loop

For a circular loop of radius R carrying current I:

B = μ₀I / 2R

The direction is perpendicular to the plane of the loop.

Use the right-hand thumb rule: curl your fingers in the direction of current. Your thumb gives the direction of magnetic field.

3. Magnetic Field on the Axis of a Circular Loop

At a distance x from the centre along the axis of a circular loop:

B = μ₀IR² / [2(R²+x²)^3/2]

The field is directed along the axis of the loop.

At x = 0:

B = μ₀I / 2R

4. Two Coaxial Current-Carrying Loops

Two circular loops are called coaxial when their axes lie along the same straight line.

If both loops produce magnetic fields in the same direction:

B_net = B₁ + B₂

If they produce fields in opposite directions:

B_net = |B₁-B₂|

For coaxial loops, the fields at a common point are generally parallel or antiparallel because both fields lie along the common axis.

5. Two Coaxial Loops at a Common Centre

Suppose two loops have the same centre and common axis. Their radii are R₁ and R₂ and currents are I₁ and I₂.

Their individual fields at the centre are:

B₁ = μ₀I₁ / 2R₁

B₂ = μ₀I₂ / 2R₂

If currents produce fields in the same direction:

B = μ₀/2 (I₁/R₁ + I₂/R₂)

If opposite:

B = μ₀/2 |I₁/R₁ - I₂/R₂|

6. Two Mutually Perpendicular Loops

Suppose two loops are mutually perpendicular and have a common centre. Their magnetic fields at the centre are also mutually perpendicular.

Let:

B₁ = μ₀I₁/2R₁

and:

B₂ = μ₀I₂/2R₂

Since B₁ ⟂ B₂:

B_net = √(B₁²+B₂²)

Substituting:

B_net = (μ₀/2) √[(I₁/R₁)²+(I₂/R₂)²]

7. Direction of Resultant Field

If B₁ and B₂ are perpendicular, the angle θ made by the resultant field with B₁ is:

tanθ = B₂/B₁

Therefore:

θ = tan⁻¹(B₂/B₁)

For the special case B₁ = B₂:

θ = 45°

The resultant magnetic field lies in the plane defined by the two perpendicular field vectors.

8. Vector Analysis

Let:

B₁ = B₁ î

and:

B₂ = B₂ ĵ

Then:

B⃗ = B₁î + B₂ĵ

Magnitude:

|B⃗| = √(B₁²+B₂²)

Direction:

tanθ = B₂/B₁

9. Two Loops Making an Arbitrary Angle

If two magnetic-field vectors B₁ and B₂ make an angle φ, then:

B_net = √(B₁²+B₂²+2B₁B₂cosφ)

The direction θ of the resultant with respect to B₁ is:

tanθ = B₂sinφ / (B₁+B₂cosφ)

Special cases:

Angle φ	Resultant
0°	B₁+B₂
90°	√(B₁²+B₂²)
180°	|B₁-B₂|

10. Solved Example 1 — Equal Perpendicular Fields

Two identical loops carry equal currents. Their planes are perpendicular. If each produces a magnetic field B at the common centre, find the resultant field.

Since:

B₁=B₂=B

and:

B₁⊥B₂

Therefore:

B_net=√(B²+B²)

B_net=√2 B

Direction:

θ=tan⁻¹(1)=45°

Resultant field makes 45° with each field.

11. Solved Example 2 — Different Perpendicular Fields

At a common centre, two perpendicular loops produce fields of 3 mT and 4 mT. Find the magnitude and direction of the resultant.

B=√(3²+4²)

B=5 mT

Direction with respect to 3 mT field:

tanθ=4/3

θ≈53.1°

12. Solved Example 3 — Coaxial Loops

Two coaxial loops produce magnetic fields 6 mT and 2 mT at a point on their common axis. If both fields are in the same direction, find the net field.

B_net=B₁+B₂

B_net=8 mT

If they were opposite, the result would be:

|6-2|=4 mT

13. Solved Example 4 — Opposite Fields

Two loops at a common centre produce fields 5 mT and 8 mT along opposite directions. Find the resultant.

B_net=|B₂-B₁|

B_net=3 mT

The direction is the direction of the 8 mT field.

14. Zero Resultant Field

For two coaxial loops producing opposite fields, the net field can be zero if:

B₁=B₂

For loops at the same centre:

I₁/R₁ = I₂/R₂

Therefore:

If I₁R₂ = I₂R₁, the fields can cancel at the common centre.

15. Field on the Common Axis of Two Coaxial Loops

For loop 1:

B₁= μ₀I₁R₁² / [2(R₁²+x₁²)^3/2]

For loop 2:

B₂= μ₀I₂R₂² / [2(R₂²+x₂²)^3/2]

If both fields are in the same direction:

B_net=B₁+B₂

If opposite:

B_net=|B₁-B₂|

16. Important Formula Sheet

Situation	Formula
Field at centre	B=μ₀I/2R
Field on axis	B=μ₀IR²/[2(R²+x²)3/2]
Same direction	B=B₁+B₂
Opposite direction	B=|B₁-B₂|
Perpendicular fields	B=√(B₁²+B₂²)
Arbitrary angle φ	B=√(B₁²+B₂²+2B₁B₂cosφ)
Direction for perpendicular fields	tanθ=B₂/B₁

Section A — 1 Mark MCQs
10 Questions

Q1. Magnetic field at the centre of a circular loop is: 1 Mark

A) μ₀I/R

B) μ₀I/2R

C) μ₀IR/2

D) 2μ₀I/R

Answer: B

Q2. The magnetic field at the centre of a loop is perpendicular to: 1 Mark

A) Plane of the loop

B) Current

C) Radius only

D) Wire only

Answer: A

Q3. Two perpendicular magnetic fields B and B have resultant: 1 Mark

A) B

B) 2B

C) √2B

D) B/2

Answer: C

Q4. If B₁ and B₂ are perpendicular, the direction satisfies: 1 Mark

A) tanθ=B₁/B₂

B) tanθ=B₂/B₁

C) θ=0

D) θ=180°

Answer: B

Q5. For two fields in the same direction, the resultant is: 1 Mark

A) B₁-B₂

B) |B₁-B₂|

C) B₁+B₂

D) √(B₁²+B₂²)

Answer: C

Q6. Two equal perpendicular fields make what angle with the resultant? 1 Mark

A) 30°

B) 45°

C) 60°

D) 90°

Answer: B

Q7. The field on the axis of a circular loop decreases with distance from the loop. 1 Mark

A) True

B) False

Answer: A

Q8. For two opposite equal fields, resultant field is: 1 Mark

A) 2B

B) B

C) √2B

D) Zero

Answer: D

Q9. Coaxial loops have: 1 Mark

A) Parallel axes

B) Same axis

C) Perpendicular axes

D) No axis

Answer: B

Q10. The magnetic field at the centre of a loop is directly proportional to: 1 Mark

A) R

B) 1/R

C) R²

D) 1/R²

Answer: B

Section B — 2 Marks
10 Questions

Q11. Write the expression for magnetic field at the centre of a circular loop. 2 Marks

B=μ₀I/2R

Q12. What is meant by coaxial loops? 2 Marks

Two loops are coaxial when their axes lie along the same straight line.

Q13. Two perpendicular fields are 3 T and 4 T. Find the resultant. 2 Marks

B=√(3²+4²)=5 T

Q14. State the right-hand thumb rule. 2 Marks

If the fingers of the right hand curl in the direction of current, the thumb gives the direction of magnetic field.

Q15. What is the resultant of two opposite fields B₁ and B₂? 2 Marks

B=|B₁-B₂|

Q16. Find the direction of resultant if B₁=B₂ and they are perpendicular. 2 Marks

tanθ=B₂/B₁=1

Therefore: θ=45°.

Q17. Write the magnetic field on the axis of a circular loop. 2 Marks

B=μ₀IR²/[2(R²+x²)^3/2]

Q18. When can two coaxial loop fields cancel each other? 2 Marks

They cancel when they are opposite in direction and have equal magnitudes: B₁=B₂.

Q19. If the radius of a loop doubles while current remains constant, what happens to its central magnetic field? 2 Marks

Since B∝1/R, the magnetic field becomes half.

Q20. Why should magnetic fields be added vectorially? 2 Marks

Because magnetic field has both magnitude and direction. Therefore, superposition follows vector addition.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the resultant magnetic field of two perpendicular loops. 3 Marks

Let the fields be:

B₁=μ₀I₁/2R₁

and:

B₂=μ₀I₂/2R₂

Since:

B₁⊥B₂

Therefore:

B=√(B₁²+B₂²)

Hence:

B=(μ₀/2)√[(I₁/R₁)²+(I₂/R₂)²]

Q22. Two perpendicular loops produce magnetic fields 6 μT and 8 μT at their common centre. Find the magnitude and direction of the resultant. 3 Marks

B=√(6²+8²)=10 μT

Direction with respect to 6 μT field:

tanθ=8/6=4/3

B=10 μT, θ≈53.1°

Q23. Two coaxial loops produce fields 7 mT and 3 mT in the same direction. Find the resultant. 3 Marks

Since fields are in the same direction:

B=7+3

B=10 mT

Q24. Two coaxial loops produce 10 mT and 6 mT in opposite directions. Find the resultant and its direction. 3 Marks

B=|10-6|

B=4 mT

Direction is that of the 10 mT field.

Q25. Two identical perpendicular loops carry equal currents. If the field produced by one loop is B, find the resultant field and its direction. 3 Marks

B₁=B₂=B

B_net=√(B²+B²)=√2B

Direction:

tanθ=1

B_net=√2B, θ=45°

Q26. Two loops have radii R and 2R and carry currents I and 2I. Their planes are perpendicular. Find the resultant field at their common centre. 3 Marks

For loop 1:

B₁=μ₀I/2R

For loop 2:

B₂=μ₀(2I)/(2×2R)

Therefore:

B₂=μ₀I/2R=B₁

Since perpendicular:

B=√2B₁

B=(μ₀I/2R)√2

Q27. Two circular loops have the same centre and same radius R. Their currents are I and 3I in opposite directions. Find the net magnetic field at the centre. 3 Marks

Individual fields:

B₁=μ₀I/2R

B₂=3μ₀I/2R

Opposite directions:

B=|B₂-B₁|

B=μ₀I/R

Direction is that of the larger field.

Q28. Two magnetic fields of magnitude B₁ and B₂ make an angle 60°. Derive the magnitude of the resultant. 3 Marks

Using vector addition:

B²=B₁²+B₂²+2B₁B₂cos60°

Since:

cos60°=1/2

Therefore:

B=√(B₁²+B₂²+B₁B₂)

Q29. Two perpendicular loop fields are B and 2B. Find the magnitude and direction of the resultant with respect to B. 3 Marks

Magnitude:

B_net=√(B²+4B²)=√5B

Direction:

tanθ=2B/B=2

B_net=√5B
θ=tan⁻¹2≈63.4°

Q30. Two coaxial circular loops have the same radius R and carry currents I and 2I in the same direction. Find the magnetic field at their common centre. 3 Marks

For first loop:

B₁=μ₀I/2R

For second:

B₂=μ₀(2I)/2R

Since both fields have the same direction:

B=B₁+B₂

B=3μ₀I/2R

Final Quick Revision

B_centre=μ₀I/2R

B_axis=μ₀IR²/[2(R²+x²)^3/2]

Same direction → B=B₁+B₂

Opposite direction → B=|B₁-B₂|

Perpendicular → B=√(B₁²+B₂²)

Arbitrary angle → B=√(B₁²+B₂²+2B₁B₂cosφ)

tanθ=B₂/B₁

Competitive Exam Trick:

1. First find the magnetic field produced by each loop.
2. Use the right-hand thumb rule to determine direction.
3. Check whether the fields are parallel, opposite or perpendicular.
4. Only then apply scalar or vector addition.

For perpendicular fields, never use B₁+B₂ directly. Use √(B₁²+B₂²).

Magnetic Fields of Intersecting and Coaxial Loops
CBSE + Competitive Physics | Concept + Formula + Solved Examples + 30 Questions