Moving Charges and Magnetism: Class 12 Physics NCERT

The Magnetic Effect of Electric Current

For centuries, electricity and magnetism were considered separate phenomena. It was Hans Christian Ørsted who, in 1820, accidentally discovered that an electric current flowing through a wire could deflect a compass needle, proving a direct relationship between electricity and magnetism. This groundbreaking discovery established that moving electric charges produce magnetic fields. This is distinct from electrostatic fields, which are produced by stationary charges.

Just as stationary charges create an electric field (E⃗ ), moving charges (i.e., electric currents) create both an electric field and a magnetic field (B⃗ ). The direction of the magnetic field produced by a current can be determined by the Right-Hand Thumb Rule: if you point your right thumb in the direction of the current, your curled fingers indicate the direction of the magnetic field lines around the conductor. Understanding this fundamental principle is crucial for building a strong foundation in electromagnetism, as it leads us to quantitative laws like the Biot-Savart Law and Ampere's Circuital Law, which we will explore next to precisely describe these magnetic fields.

Lorentz Force: Force on a Charge in Magnetic Field

Magnetic Lorentz Force

The force experienced by a moving charge (q) in a magnetic field (B⃗ ) is called the magnetic Lorentz force. It is given by the vector cross product: $\vec{F}_m = q(\vec{v} \times \vec{B})$, where $\vec{v}$ is the velocity of the charge. The direction of this force is perpendicular to both the velocity vector and the magnetic field vector. Its magnitude is $F_m = qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. This force is maximum when the charge moves perpendicular to the field ($\theta = 90^\circ$) and zero when it moves parallel or anti-parallel ($\theta = 0^\circ$ or $180^\circ$) to the field.

Total Lorentz Force

When a charged particle moves in a region where both electric ($\vec{E}$) and magnetic ($\vec{B}$) fields are present, the total force experienced by the charge (q) is the sum of the electric force and the magnetic force. This is known as the Total Lorentz Force: $\vec{F} = \vec{F}_e + \vec{F}_m = q\vec{E} + q(\vec{v} \times \vec{B})$. This comprehensive force law is central to understanding the motion of charged particles in electromagnetic fields, impacting applications like mass spectrometers and particle accelerators.

Fleming's Left-Hand Rule

This rule helps determine the direction of the force on a current-carrying conductor placed in a magnetic field, or equivalently, the force on a moving positive charge. If you extend the forefinger, central finger, and thumb of your left hand mutually perpendicular to each other, such that the forefinger points in the direction of the magnetic field ($\vec{B}$), the central finger points in the direction of the current (or velocity of positive charge, $\vec{v}$), then the thumb will point in the direction of the force ($\vec{F}$) experienced by the conductor/charge.

Biot-Savart Law: Calculating Magnetic Fields

1. Understand the Principle — The Biot-Savart Law is a fundamental law in magnetostatics that describes the magnetic field generated by a steady electric current. It states that the magnetic field $\vec{dB}$ at a point P due to a current element $I\vec{dl}$ is directly proportional to the current I, the length $dl$ of the element, and the sine of the angle $\theta$ between the current element and the position vector $\vec{r}$ (from the element to point P), and inversely proportional to the square of the distance r. It's the magnetic analogue to Coulomb's Law for electric fields.
2. Learn the Mathematical Form — The law is expressed in vector form as: $\vec{dB} = \frac{\mu_0}{4\pi} \frac{I(\vec{dl} \times \vec{r})}{r^3}$. Here, $\mu_0$ is the permeability of free space (a constant, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$), I is the current, $\vec{dl}$ is a vector element of the wire in the direction of current, $\vec{r}$ is the position vector from $dl$ to the observation point, and r is the magnitude of $\vec{r}$. The cross product $(\vec{dl} \times \vec{r})$ dictates the direction of $\vec{dB}$ using the Right-Hand Rule.
3. Apply to Common Geometries — To find the total magnetic field $\vec{B}$ due to a finite current distribution, you need to integrate $\vec{dB}$ over the entire length of the conductor: $\vec{B} = \int \vec{dB}$. This integration can be complex. Common applications include calculating the magnetic field due to a long straight wire, a circular current loop at its center, or along its axis. For instance, the magnetic field at the center of a circular loop of radius R carrying current I is $B = \frac{\mu_0 I}{2R}$.
4. Direction using Right-Hand Rule — For a current element $\vec{dl}$, point your right thumb in the direction of $\vec{dl}$, and curl your fingers. The direction in which your fingers curl at point P gives the direction of $\vec{dB}$. Alternatively, for the cross product $(\vec{dl} \times \vec{r})$, use the standard right-hand rule for cross products: point fingers in direction of $\vec{dl}$, curl towards $\vec{r}$, thumb gives direction of $\vec{dB}$.

Ampere's Circuital Law and its Applications

• Ampere's Circuital Law provides an alternative, often simpler, method to calculate magnetic fields, especially for situations with high symmetry. It states that the line integral of the magnetic field $\vec{B}$ around any closed loop (called an Amperian loop) is equal to $\mu_0$ times the total current (I_enclosed) passing through the area enclosed by the loop. Mathematically: $\oint \vec{B} \cdot \vec{dl} = \mu_0 I_{\text{enclosed}}$. This law is analogous to Gauss's Law in electrostatics. Example: Magnetic Field due to a Long Straight Current-Carrying Wire Let's apply Ampere's Law to find the magnetic field B at a distance 'r' from a long straight wire carrying a steady current 'I'. Step 1: Choose an Amperian Loop. Due to the cylindrical symmetry of the wire, the magnetic field lines will be concentric circles around the wire. We choose a circular Amperian loop of radius 'r' centered on the wire and lying in a plane perpendicular to the wire. Step 2: Evaluate the line integral. Along this circular loop, the magnetic field $\vec{B}$ is tangential to the loop at every point and has the same magnitude 'B' due to symmetry. Also, the direction of the line element $\vec{dl}$ is always tangential to the loop and in the same direction as $\vec{B}$. Therefore, $\vec{B} \cdot \vec{dl} = B dl \cos(0^\circ) = B dl$. The line integral becomes $\oint \vec{B} \cdot \vec{dl} = \oint B dl = B \oint dl$. The integral $\oint dl$ is the circumference of the circular loop, which is $2\pi r$. So, $\oint \vec{B} \cdot \vec{dl} = B (2\pi r)$. Step 3: Determine the enclosed current. The total current enclosed by our Amperian loop is simply the current 'I' flowing through the wire. Step 4: Apply Ampere's Law. According to Ampere's Law: $\oint \vec{B} \cdot \vec{dl} = \mu_0 I_{\text{enclosed}}$. Substituting our results: $B (2\pi r) = \mu_0 I$. Step 5: Solve for B. $B = \frac{\mu_0 I}{2\pi r}$. This formula gives the magnitude of the magnetic field at a distance 'r' from a long straight current-carrying wire. The direction is given by the Right-Hand Thumb Rule (concentric circles around the wire). This example beautifully demonstrates the power of Ampere's Law for symmetric situations.

Force Between Parallel Current-Carrying Conductors and Torque on a Current Loop

Understanding how magnetic fields interact with currents leads us to two significant phenomena: forces between parallel conductors and torque on current loops.

Force between Two Parallel Current-Carrying Conductors:
Consider two long, straight parallel conductors separated by a distance 'd', carrying currents $I_1$ and $I_2$ respectively. Each conductor produces a magnetic field, and each conductor then experiences a force due to the magnetic field produced by the other. Using Ampere's Law and the Lorentz force formula, it can be shown that the force per unit length between the two conductors is given by: $F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$.

Crucially, the nature of this force depends on the direction of currents:

• Parallel Currents: If the currents flow in the same direction, the force between the conductors is attractive.
• Anti-parallel Currents: If the currents flow in opposite directions, the force between the conductors is repulsive.

This interaction is so fundamental that it is used to define the SI unit of current, the Ampere. One Ampere is defined as that steady current which, when maintained in two very long, straight, parallel conductors of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to $2 \times 10^{-7}$ newtons per metre of length.

Torque on a Current Loop in a Uniform Magnetic Field:
When a current-carrying loop is placed in a uniform magnetic field, it experiences a torque. This is the principle behind electric motors. For a rectangular loop of area A carrying current I, placed in a uniform magnetic field $\vec{B}$, the torque $\vec{\tau}$ is given by: $\vec{\tau} = \vec{M} \times \vec{B}$, where $\vec{M}$ is the magnetic dipole moment of the loop. The magnetic dipole moment is defined as $\vec{M} = NI\vec{A}$, where N is the number of turns in the coil, I is the current, and $\vec{A}$ is the area vector of the loop (its magnitude is the area, and its direction is perpendicular to the loop, given by the Right-Hand Rule of curling fingers in current direction, thumb gives area vector direction).
The magnitude of the torque is $\tau = NIAB\sin\theta$, where $\theta$ is the angle between the normal to the plane of the loop and the magnetic field $\vec{B}$. The torque tends to align the magnetic dipole moment with the magnetic field.

Exam Tips and Common Pitfalls

Navigating 'Moving Charges and Magnetism' requires precision, especially with vector directions. Here are key tips and common mistakes to avoid:

1. Directional Rules are Key: Always identify whether to use the Right-Hand Thumb Rule (for magnetic field due to current), Fleming's Left-Hand Rule (for force on current/charge in B-field), or the Right-Hand Rule for cross products (e.g., for Biot-Savart Law or Lorentz force). A common mistake is interchanging these rules or applying them incorrectly.
2. Vector Cross Products: Remember that $\vec{A} \times \vec{B}$ is perpendicular to both $\vec{A}$ and $\vec{B}$. The order matters: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$. Be precise with the angles in $F = qvB\sin\theta$ and $\tau = NIAB\sin\theta$. Ensure $\theta$ is the angle between the correct pair of vectors (e.g., $\vec{v}$ and $\vec{B}$, or $\vec{M}$ and $\vec{B}$). For Lorentz force, force is zero if $\vec{v}$ is parallel or anti-parallel to $\vec{B}$.
3. Amperian Loop Selection: When using Ampere's Circuital Law, choose an Amperian loop that exploits the symmetry of the current distribution. The magnetic field $\vec{B}$ should either be tangential to the loop (and constant in magnitude) or perpendicular to the loop (resulting in $\vec{B} \cdot \vec{dl} = 0$). Incorrect loop selection makes the integration complex or impossible.
4. Permeability Constant: Don't forget the $\mu_0/(4\pi)$ term in Biot-Savart Law and $\mu_0$ in Ampere's Law. Remember $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
5. Units: Always include correct SI units in your final answers. Magnetic field is in Tesla (T), current in Ampere (A), length in meter (m), force in Newton (N), charge in Coulomb (C), velocity in m/s.
6. Conceptual Understanding: Don't just memorize formulas. Understand why current-carrying wires attract/repel, or why a loop experiences torque. This conceptual clarity helps in solving application-based problems.

Worked Examples

• Example 1: Force on a Proton in a Magnetic Field A proton is moving horizontally towards the East with a velocity of $4 \times 10^6 \text{ m/s}$. It enters a uniform magnetic field of $0.5 \text{ T}$ directed vertically upwards. Calculate the magnetic force acting on the proton. (Charge of proton $q = 1.6 \times 10^{-19} \text{ C}$). Step 1: Identify given quantities and directions. Velocity of proton $\vec{v} = 4 \times 10^6 \text{ m/s}$ (East). Magnetic field $\vec{B} = 0.5 \text{ T}$ (Vertically upwards). Charge of proton $q = 1.6 \times 10^{-19} \text{ C}$. Step 2: Apply the Lorentz force formula. The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$. Magnitude: $F_m = qvB\sin\theta$. Here, the velocity (East) and magnetic field (upwards) are perpendicular to each other, so $\theta = 90^\circ$ and $\sin\theta = 1$. Step 3: Calculate the magnitude of the force. $F_m = (1.6 \times 10^{-19} \text{ C}) \times (4 \times 10^6 \text{ m/s}) \times (0.5 \text{ T}) \times 1$ $F_m = 1.6 \times 4 \times 0.5 \times 10^{-19+6} \text{ N}$ $F_m = 3.2 \times 10^{-13} \text{ N}$. Step 4: Determine the direction of the force using Fleming's Left-Hand Rule or Right-Hand Rule for cross product. Using Fleming's Left-Hand Rule: Forefinger (B) points upwards, Central finger (v) points East. The Thumb will point towards North. Using Right-Hand Rule for $\vec{v} \times \vec{B}$: Point fingers East (v), curl them upwards (B). Your thumb will point North. Final Answer: The magnetic force acting on the proton is $3.2 \times 10^{-13} \text{ N}$ directed towards the North. Example 2: Magnetic Field at the Center of a Current Loop A circular coil of radius 10 cm has 50 turns and carries a current of 2 A. Calculate the magnetic field at its center. Step 1: Identify given quantities and relevant formula. Radius of coil, $R = 10 \text{ cm} = 0.1 \text{ m}$. Number of turns, $N = 50$. Current, $I = 2 \text{ A}$. Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$. Formula for magnetic field at the center of a circular coil: $B = \frac{\mu_0 N I}{2R}$. Step 2: Substitute the values into the formula. $B = \frac{(4\pi \times 10^{-7} \text{ T m/A}) \times 50 \times (2 \text{ A})}{2 \times (0.1 \text{ m})}$ Step 3: Perform the calculation. $B = \frac{4\pi \times 10^{-7} \times 100}{0.2}$ $B = \frac{4\pi \times 10^{-7} \times 1000}{2}$ $B = 2\pi \times 10^{-4} \text{ T}$ $B \approx 2 \times 3.14159 \times 10^{-4} \text{ T}$ $B \approx 6.283 \times 10^{-4} \text{ T}$. Final Answer: The magnetic field at the center of the coil is approximately $6.283 \times 10^{-4} \text{ T}$. The direction would be perpendicular to the plane of the coil, determined by the Right-Hand Thumb Rule (curling fingers in direction of current, thumb points to B).

Practice Questions with Solutions

• Q: An electron (charge $e = -1.6 \times 10^{-19} \text{ C}$, mass $m = 9.1 \times 10^{-31} \text{ kg}$) is projected with a velocity $v = 2 \times 10^7 \text{ m/s}$ along the positive x-axis. It enters a uniform magnetic field $B = 0.2 \text{ T}$ directed along the positive y-axis. Describe the path of the electron and calculate the magnetic force acting on it. A: Step 1: Identify given values and formula. $q = -1.6 \times 10^{-19} \text{ C}$, $v = 2 \times 10^7 \text{ m/s}$ (along +x), $B = 0.2 \text{ T}$ (along +y). The force formula is $\vec{F} = q(\vec{v} \times \vec{B})$. Step 2: Calculate the magnitude of the force. Since $\vec{v}$ is along +x and $\vec{B}$ is along +y, they are perpendicular, so $\theta = 90^\circ$. $F = |q|vB\sin\theta = (1.6 \times 10^{-19}) \times (2 \times 10^7) \times 0.2 \times \sin(90^\circ) = 6.4 \times 10^{-13} \text{ N}$. Step 3: Determine the direction of the force. For $\vec{v} \times \vec{B}$, (x-direction $\times$ y-direction) gives z-direction. However, the charge is negative (electron), so the force direction will be opposite to the result of $\vec{v} \times \vec{B}$. So, the force is along the negative z-axis. This force is perpendicular to both velocity and magnetic field. Step 4: Describe the path. Since the magnetic force is always perpendicular to the velocity, it provides the necessary centripetal force. The electron will follow a circular path in the xz-plane. Final answer: The magnetic force is $6.4 \times 10^{-13} \text{ N}$ along the negative z-axis. The electron will follow a circular path in the xz-plane.
• Q: A long straight wire carries a current of $5 \text{ A}$. Calculate the magnitude of the magnetic field at a point $5 \text{ cm}$ away from the wire. A: Step 1: Identify given values. Current $I = 5 \text{ A}$. Distance $r = 5 \text{ cm} = 0.05 \text{ m}$. Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$. Step 2: Use the formula for magnetic field due to a long straight wire: $B = \frac{\mu_0 I}{2\pi r}$. Step 3: Substitute the values and calculate. $B = \frac{(4\pi \times 10^{-7} \text{ T m/A}) \times 5 \text{ A}}{2\pi \times 0.05 \text{ m}} = \frac{2 \times 10^{-7} \times 5}{0.05} = \frac{10 \times 10^{-7}}{0.05} = \frac{1000 \times 10^{-7}}{5} = 200 \times 10^{-7} = 2 \times 10^{-5} \text{ T}$. Final answer: The magnetic field at $5 \text{ cm}$ from the wire is $2 \times 10^{-5} \text{ T}$.
• Q: Two long and parallel straight wires are placed $10 \text{ cm}$ apart in air. If they carry currents of $4 \text{ A}$ and $6 \text{ A}$ respectively in the same direction, calculate the force per unit length between them. What is the nature of this force? A: Step 1: Identify given values. Currents $I_1 = 4 \text{ A}$, $I_2 = 6 \text{ A}$. Distance $d = 10 \text{ cm} = 0.1 \text{ m}$. $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$. Step 2: Use the formula for force per unit length between parallel wires: $F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$. Step 3: Substitute values and calculate. $F/L = \frac{(4\pi \times 10^{-7}) \times 4 \times 6}{2\pi \times 0.1} = \frac{2 \times 10^{-7} \times 24}{0.1} = \frac{48 \times 10^{-7}}{0.1} = 480 \times 10^{-7} = 4.8 \times 10^{-5} \text{ N/m}$. Step 4: Determine the nature of the force. Since the currents are in the same direction, the force is attractive. Final answer: The force per unit length between the wires is $4.8 \times 10^{-5} \text{ N/m}$, and it is attractive.
• Q: A circular coil of 20 turns and radius $10 \text{ cm}$ is placed in a uniform magnetic field of $0.10 \text{ T}$ normally to the plane of the coil. If the current in the coil is $5 \text{ A}$, calculate the torque acting on the coil. A: Step 1: Identify given values. Number of turns $N = 20$. Radius $R = 10 \text{ cm} = 0.1 \text{ m}$. Magnetic field $B = 0.10 \text{ T}$. Current $I = 5 \text{ A}$. Step 2: Calculate the area of the coil. $A = \pi R^2 = \pi (0.1)^2 = 0.01\pi \text{ m}^2$. Step 3: Determine the angle $\theta$. The coil is placed normally to the plane of the coil, meaning the normal to the coil's plane (which is the direction of the magnetic moment $\vec{M}$) is perpendicular to the magnetic field. Wait, "normally to the plane of the coil" means the magnetic field lines are perpendicular to the plane of the coil, which means the normal to the coil's plane is parallel to the magnetic field. So, $\theta = 0^\circ$. Step 4: Apply the torque formula: $\tau = NIAB\sin\theta$. Since $\theta = 0^\circ$, $\sin\theta = 0$. Final answer: The torque acting on the coil is $0 \text{ N m}$ because the magnetic field is aligned with the magnetic dipole moment of the coil.

Frequently Asked Questions