Introduction | The Oersted Revolution | Class 12 Physics Chapter 4 Moving Charges and Magnetism Notes

Students, imagine a world where electricity and magnetism are completely separate things. That was the world of Physics before 1820. Then, a Danish physicist named Hans Christian Oersted made an accidental discovery that changed everything.

He noticed that a simple compass needle twitched when placed near a wire carrying current. This observation proved that electricity and magnetism are not separate. Moving charges (current) produce magnetic fields. This connection is the heartbeat of our modern world—from the fans cooling this room to the motors in electric cars. In this chapter, we will explore exactly how currents create magnetic fields and how magnetic fields push moving charges.

1. Magnetic Force (Lorentz Force)

1.1 Force on a Moving Charge

We know that a stationary charge creates an electric field. But does a magnetic field affect a stationary charge? No. A magnetic field is like a sleeping giant—it only wakes up and exerts force when a charge is moving.

If a charge $q$ moves with velocity $v$ in a magnetic field $B$, the magnetic force is given by the vector cross product:
$$F_m = q(v \times B)$$
Magnitude: $$F_m = qvB \sin\theta$$

Teacher’s Insight:

• Stationary Charge ($v=0$): Force is Zero. Magnetic fields ignore charges at rest.
• Parallel Motion ($\theta = 0^\circ$): Force is Zero. If you run parallel to the field lines, you are safe.
• Perpendicular Motion ($\theta = 90^\circ$): Force is Maximum ($F = qvB$).

Figure-1: The Right-Hand Rule. Point fingers in direction of velocity ($v$), curl them towards Magnetic Field ($B$), and your thumb points to the Force ($F$).

1.2 Combined Force (Lorentz Force)

If a particle moves in a region where both electric and magnetic fields exist, the total force is simply the vector sum:
$$F_{total} = F_{electric} + F_{magnetic} = qE + q(v \times B)$$

2. Motion in a Magnetic Field

This is one of the most conceptual parts of the chapter. Since the magnetic force is always perpendicular to velocity, it acts exactly like a Centripetal Force.

Key Concept: Magnetic force changes the direction of the particle but never its speed or Kinetic Energy. Work done by magnetic force is always zero!

Case A: Circular Motion ($v \perp B$)

When velocity is perpendicular to the field, the particle goes in a perfect circle.
$$Centripetal \, Force = Magnetic \, Force$$
$$\frac{mv^2}{r} = qvB \implies r = \frac{mv}{qB}$$

Cyclotron Frequency:
The time to complete one circle ($T$) is fascinating because it does not depend on speed ($v$) or radius ($r$).
$$T = \frac{2\pi m}{qB}$$
Frequency $\nu = \frac{qB}{2\pi m}$

Case B: Helical Motion

If the particle enters at an angle, velocity splits into two components:

• $v_{\parallel}$ (Parallel to B): Pulls the particle forward in a straight line.
• $v_{\perp}$ (Perpendicular to B): Makes the particle rotate in a circle.

The result is a spiral path called a Helix.

Figure-2: Helical motion. The ‘Pitch’ is the horizontal distance covered in one full rotation.

3. Force on a Current-Carrying Wire

If a single moving charge feels a force, a wire full of flowing charges (current) must also feel a force. This is the principle behind Electric Motors.

$$F = I (l \times B)$$
Magnitude: $$F = IlB \sin\theta$$
Where $l$ is the length of the conductor inside the field.

4. Calculating Magnetic Fields (Biot-Savart Law)

Up until now, we discussed the force on a charge. Now, let’s ask: How strong is the magnetic field created by a current?

Biot-Savart Law: For a tiny element of wire $dl$ carrying current $I$, the magnetic field $dB$ at distance $r$ is:
$$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$$
Here, $\mu_0 = 4\pi \times 10^{-7} Tm/A$ is the permeability of free space.

Application: Field at the Center of a Circular Coil

Using the law above, for a circular loop of radius $R$ and $N$ turns, the field at the center is:
$$B = \frac{\mu_0 N I}{2R}$$

Teacher’s Tip: Notice the difference? For a straight wire, field $\propto 1/r$. For a coil center, field $\propto 1/R$. Don’t confuse the formulas!

5. Ampere’s Circuital Law

Just like Gauss’s Law made Electrostatics easy, Ampere’s Law makes Magnetism easy for symmetrical problems.

Statement: The line integral of magnetic field around any closed loop is equal to $\mu_0$ times the net current passing through the loop.
$$\oint B \cdot dl = \mu_0 I_{enclosed}$$

Key Derivations using Ampere’s Law:

1. Infinite Straight Wire: Field at distance $r$:
   $$B = \frac{\mu_0 I}{2\pi r}$$
   (Field lines are concentric circles).
2. Solenoid: A long coil with many turns. The field inside is uniform and strong, while outside is nearly zero.
   $$B = \mu_0 n I$$
   (Where $n$ is turns per unit length, $N/L$).

Figure-3: Magnetic field lines inside a solenoid are parallel, indicating a uniform field.

6. Force Between Two Parallel Wires

This leads to a very famous definition. If two wires carry current:

• Same Direction: They ATTRACT each other.
• Opposite Direction: They REPEL each other.

The force per unit length is:
$$f = \frac{\mu_0 I_1 I_2}{2\pi d}$$

Definition of 1 Ampere: One Ampere is that current which, when flowing in two parallel wires 1 meter apart in vacuum, produces a force of $2 \times 10^{-7} N$ per meter of length.

7. Torque and The Galvanometer

7.1 Torque on a Rectangular Loop

A current loop in a magnetic field feels NO net force, but it feels a Torque ($\tau$). This twist is what spins the fan above your head.
$$\tau = m \times B = NIAB \sin\theta$$
(Here $\theta$ is the angle between the normal to the coil and the field).

7.2 The Moving Coil Galvanometer

This is a sensitive device to detect current.

• Principle: Current in the coil creates a torque, which twists a suspension spring.
• Radial Field: We use curved magnets and a soft iron core to ensure the field is always radial. This makes the torque constant and maximum ($\theta=90^\circ$).

Figure-4: The Moving Coil Galvanometer. The deflection $\phi$ is directly proportional to current $I$.

7.3 Conversion to Ammeter and Voltmeter

A galvanometer is too sensitive to be used directly. We modify it:

• To Ammeter: Connect a very low resistance (Shunt) in Parallel.
• To Voltmeter: Connect a very high resistance in Series.

Solved Numericals (Teacher’s Walkthrough)

Numerical 1 (Circular Path):

Question: An electron enters a 0.2 T magnetic field at a speed of $3 \times 10^7$ m/s perpendicular to the field. Find the radius of its path.

Solution:
We know, $r = \frac{mv}{qB}$
Mass of electron $m = 9.1 \times 10^{-31}$ kg.
$$r = \frac{(9.1 \times 10^{-31}) \times (3 \times 10^7)}{(1.6 \times 10^{-19}) \times 0.2}$$
$$r = \frac{27.3 \times 10^{-24}}{0.32 \times 10^{-19}} \approx 8.5 \times 10^{-4} \text{ m} = 0.85 \text{ mm}$$

Numerical 2 (Solenoid):

Question: A 0.5 m long solenoid has 500 turns and carries 5 A current. What is the magnetic field inside?

Solution:
First, find $n$ (turns per unit length): $n = \frac{500}{0.5} = 1000$ turns/m.
Formula: $B = \mu_0 n I$
$$B = (4\pi \times 10^{-7}) \times 1000 \times 5$$
$$B = 20\pi \times 10^{-4} \approx 6.28 \times 10^{-3} \text{ T}$$

Practice Set with Complete Solutions

A. Very Short Answer (VSA) (1 Mark Each)

Q1. Under what condition does a moving charge experience zero magnetic force?

Answer: A moving charge experiences zero magnetic force when it moves parallel ($0^{\circ}$) or anti-parallel ($180^{\circ}$) to the direction of the magnetic field.

Reasoning: Formula is $F = qvB \sin\theta$. If $\theta = 0^{\circ}$, then $\sin\theta = 0$.

Q2. Does the kinetic energy of a charged particle change when moving in a magnetic field?

Answer: No, the kinetic energy remains constant.

Reasoning: The magnetic force is always perpendicular to the velocity ($F \perp v$). Therefore, the work done by the magnetic force is zero ($W = F \cdot s = 0$). If Work Done is zero, there is no change in Kinetic Energy.

B. Short Answer (SA) (2-3 Marks Each)

Q3. Why is a Soft Iron core used in a Moving Coil Galvanometer?

Answer: It serves two main purposes:

1. Increases Field Strength: Soft iron is ferromagnetic and concentrates magnetic field lines, making the field stronger.

2. Radial Field: It helps make the field radial, ensuring the angle between the coil area vector and magnetic field is always $90^{\circ}$, which keeps torque maximum.

Q4. Two parallel wires carry current in the same direction. Do they attract or repel? Explain.

Answer: They Attract each other.

Explanation: Wire 1 creates an inward magnetic field at the position of Wire 2. By Fleming’s Left-Hand Rule, the force on Wire 2 points towards Wire 1.

Q5. An electron and a proton enter a magnetic field with the same velocity. Which one describes a smaller circle?

Answer: The Electron moves in a smaller circle.

Reasoning: Radius $r = \frac{mv}{qB}$. Since $v$, $q$, and $B$ are the same, $r \propto m$. The mass of an electron ($9.1 \times 10^{-31}$ kg) is much smaller than a proton, so its radius is smaller.

C. Long Answer (LA) (5 Marks Each)

Q6. Using Biot-Savart Law, derive the expression for the magnetic field on the axis of a circular current loop.

Answer:

1. Consider a loop of radius $R$ with current $I$. Take a small element $dl$.

2. According to Biot-Savart Law: $dB = \frac{\mu_0 I dl \sin 90^{\circ}}{4\pi r^2}$.

3. Resolve $dB$ into vertical ($dB \cos\theta$) and horizontal ($dB \sin\theta$) components. Vertical components cancel out.

4. Integrate the horizontal components: $B = \int dB \sin\theta$.

5. Substituting $\sin\theta = R/r$ and integrating $\int dl = 2\pi R$:

$$B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$$

Q7. Describe the construction and working of a Moving Coil Galvanometer.

Answer:

Construction: It consists of a coil suspended between concave pole pieces of a strong magnet. A soft iron core is placed inside to make the field radial.

Principle: A current-carrying coil in a magnetic field experiences a torque $\tau = NIAB \sin 90^{\circ} = NIAB$.

Working: This magnetic torque is balanced by the restoring torque of the spring ($k\phi$).

$$NIAB = k\phi \implies \phi = \left(\frac{NAB}{k}\right) I$$

Thus, deflection $\phi$ is directly proportional to current $I$.

D. Case-Based & Assertion-Reason

Q8. The Cyclotron: A device uses magnetic fields to make particles go in circles and electric fields to speed them up.

(i) What happens to the radius as the particle gains speed?

Answer: The radius increases. (From $r = mv/qB$, $r \propto v$).

(ii) Why can’t we accelerate neutrons using a cyclotron?

Answer: Cyclotrons use electric and magnetic forces to accelerate particles. Since neutrons are neutral (charge $q=0$), they experience no force ($F=0$) and cannot be accelerated.

Q9. Assertion (A): The magnetic field at the ends of a very long current-carrying solenoid is half of that at the center.

Reason (R): The field outside a solenoid is zero.

Answer: Both are true, but R is not the correct explanation for A. The calculation shows $B_{end} = \frac{1}{2} \mu_0 n I$ while $B_{center} = \mu_0 n I$.