Chapter 4: Moving Charges and Magnetism

1. Magnetic Field Due to a Current-Carrying Conductor

1.1 Biot-Savart Law

• Formula: $d\mathbf{B}=\frac{{\mu }_0}{4\pi }\cdot \frac{I,d\mathbf{l}\times \hat{\mathbf{r}}}{r^2}$
• Explanation: This law provides the magnetic field $d\mathbf{B}$ at a point due to an infinitesimal current element $I,d\mathbf{l}$.
• Conditions: The formula applies to steady currents and is used to calculate magnetic fields in configurations like circular loops and straight conductors.

1.2 Magnetic Field Due to a Straight Current-Carrying Conductor

• Formula: $B=\frac{{\mu }_{0}I}{2\pi r}$
• Explanation: The magnetic field $B$ at a distance $r$ from a long, straight conductor carrying current $I$.
• Application Example: A wire carrying 10 A current produces a magnetic field of $2\times 10^{-5}$ T at a point 1 m away.
• Common Mistake: Ensure that $r$ is the perpendicular distance from the point to the conductor, not along the length of the wire.

1.3 Magnetic Field on the Axis of a Circular Current Loop

• Formula: $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$
• Explanation: The magnetic field at a point on the axis of a circular loop of radius $R$ carrying current $I$, at a distance $X$ from the center.
• Derivation: Derived using Biot-Savart law by integrating the contributions from all current elements around the loop.

2. Force on a Moving Charge in a Magnetic Field

2.1 Lorentz Force

• Formula: $\mathbf{F}=q(\mathbf{v}\times \mathbf{B})$
• Explanation: The force $\mathbf{F}$ on a charge $q$ moving with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$.
• Application Example: For a proton moving at $2\times 10^6,\text{m/s}$ perpendicular to a magnetic field of $0.1,\text{T}$, the force is $3.2\times 10^{-14},\text{N}$.
• Common Mistake: Remember that the direction of the force is given by the right-hand rule.

2.2 Force Between Two Parallel Currents

• Formula: $F/l=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$
• Explanation: The force per unit length between two parallel current-carrying wires separated by distance $d$.
• Real-life Application: This principle underlies the definition of the ampere, the SI unit of current.

3. Magnetic Moment of a Current Loop

• Formula: $\mathbf{m}=I\mathbf{A}$
• Explanation: The magnetic moment $\mathbf{m}$ of a current loop, where $I$ is the current and $\mathbf{A}$ is the area vector.
• Common Misconception: Magnetic moment is a vector quantity, and its direction is given by the right-hand thumb rule.

Chapter 5: Magnetism and Matter

1. Magnetic Field Due to a Bar Magnet

1.1 Magnetic Field on the Axis of a Bar Magnet

• Formula: $B_{\text{axial}}=\frac{{\mu }_0}{4\pi }\cdot \frac{2m}{r^3}$
• Explanation: The magnetic field at a distance $r$ from the center of a bar magnet on its axial line, where $m$ is the magnetic moment.
• Application Example: Calculate the field at 10 cm from a bar magnet with a magnetic moment of $1,{\text{A m}}^2$.
• Common Mistake: Ensure that $r$ is sufficiently large compared to the size of the magnet to use this formula.

1.2 Magnetic Field at the Equatorial Line of a Bar Magnet

• Formula: $B_{\text{equatorial}}=\frac{{\mu }_0}{4\pi }\cdot \frac{m}{r^3}$
• Explanation: The magnetic field at a distance $r$ from the center of a bar magnet on its equatorial line.

2. Gauss’s Law for Magnetism

• Statement: The net magnetic flux through any closed surface is zero.
• Formula: $\oint \mathbf{B}\cdot d\mathbf{A}=0$
• Explanation: This law implies that magnetic monopoles do not exist; the magnetic field lines are continuous and form closed loops.

3. Magnetic Properties of Materials

3.1 Magnetization and Magnetic Intensity

• Formula: $\mathbf{M}=\frac{{\mathbf{m}}_{\text{net}}}{V}$, $\mathbf{H}=\frac{1}{{\mu }_0}(\mathbf{B}-{\mu }_0\mathbf{M})$
• Explanation: Magnetization $\mathbf{M}$ is the net magnetic moment per unit volume, and magnetic intensity $\mathbf{H}$ is related to the external magnetic field $\mathbf{B}$ and magnetization $\mathbf{M}$.

3.2 Magnetic Susceptibility and Permeability

• Formula: ${\chi }_m=\frac{M}{H}$, $\mu ={\mu }_0(1+{\chi }_m)$
• Explanation: Magnetic susceptibility ${\chi }_m$ describes how much a material will become magnetized in an applied magnetic field, and permeability $\mu$ is the measure of the ability of the material to support the formation of a magnetic field.

Practice Questions

1. Calculate the magnetic field at a point 1 m away from a long straight conductor carrying 5 A current.
2. • Solution: $B=\frac{{\mu }_{0}I}{2\pi r}=\frac{4\pi \times 10^{-7}\times 5}{2\pi \times 1}=10^{-6},\text{T}$.
2. A proton moves perpendicular to a magnetic field of $0.2,\text{T}$ with a velocity of $1\times 10^7,\text{m/s}$. Calculate the force acting on it.
3. • Solution: $F=qvB=1.6\times 10^{-19}\times 1\times 10^7\times 0.2=3.2\times 10^{-13},\text{N}$.
3. Derive the expression for the magnetic field at the center of a circular loop carrying current $I$ and of radius $R$.
4. A bar magnet with a magnetic moment of $2,{\text{A m}}^2$ is placed in a uniform magnetic field of $0.5,\text{T}$. Calculate the torque acting on it when the magnet makes an angle of $30^{\circ }$ with the field.
5. • Solution: $\tau =mB\sin \theta =2\times 0.5\times \sin 30^{\circ }=0.5,\text{N m}$.

Quick Recap

• Biot-Savart Law: Fundamental to calculating magnetic fields due to current elements.
• Lorentz Force: Crucial for understanding the motion of charges in a magnetic field.
• Magnetic Properties: Understanding different types of magnetism (diamagnetism, paramagnetism, ferromagnetism) is vital for material classification.

Concept Connection

• Electromagnetism in Chemistry: The concepts of magnetic fields and forces are linked to electron configurations and chemical bonding.
• Biology: The principle of magnetic fields is used in MRI scans, illustrating the connection between physics and medical imaging.

This summary captures the critical formulae, explanations, and common pitfalls from the chapters "Moving Charges and Magnetism" and "Magnetism and Matter" from the NCERT textbook. The practice questions and derivations are tailored to enhance problem-solving skills for NEET preparation.