Introduction
In physics, the study of moving charges and magnetism is crucial for understanding the interplay between electric currents and magnetic fields. This topic is pivotal not only for NEET but also for a comprehensive understanding of electromagnetism. The phenomena can be explained using several fundamental laws and principles, which are essential for solving problems in this domain.
Magnetic Field and Its Properties
Magnetic Field ($\mathbf{B}$)
A magnetic field is a vector field that exerts a force on moving charges and magnetic dipoles. It is denoted by the symbol $\mathbf{B}$ and measured in Tesla (T).
Magnetic Force on a Moving Charge
When a charged particle with charge $q$ moves with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$, it experiences a force given by:
$$ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) $$
where $\mathbf{F}$ is the magnetic force, and $\times$ denotes the cross product.
NoteThe direction of the magnetic force is perpendicular to both the velocity of the charge and the magnetic field, as determined by the right-hand rule.
Motion of Charged Particles in Magnetic Fields
Circular Motion
A charged particle moving perpendicular to a uniform magnetic field will undergo circular motion due to the centripetal force provided by the magnetic force. The radius of the circular path, known as the Larmor radius, is given by:
$$ r = \frac{mv}{qB} $$
where $m$ is the mass of the particle, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength.
ExampleConsider a proton ($q = 1.6 \times 10^{-19}$ C, $m = 1.67 \times 10^{-27}$ kg) moving with a velocity of $10^6$ m/s in a magnetic field of $1$ T. The radius of its circular path is: $$ r = \frac{(1.67 \times 10^{-27} \text{ kg})(10^6 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C})(1 \text{ T})} \approx 0.01 \text{ m} $$
Helical Motion
If the velocity of the charged particle has a component parallel to the magnetic field, the particle will follow a helical path. The pitch of the helix, or the distance between consecutive turns, is given by:
$$ p = v_{\parallel} \cdot T $$
where $v_{\parallel}$ is the component of velocity parallel to the magnetic field and $T$ is the period of the circular motion.
Biot-Savart Law
The Biot-Savart Law relates the magnetic field generated by a current-carrying conductor to the current and the geometry of the conductor. It is given by:
$$ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{r}}{r^3} $$
where $d\mathbf{B}$ is the infinitesimal magnetic field, $\mu_0$ is the permeability of free space, $I$ is the current, $d\mathbf{l}$ is the infinitesimal length element of the conductor, and $\mathbf{r}$ is the position vector from the element to the point where the field is being calculated.
TipUse the right-hand rule to determine the direction of the magnetic field: Point the thumb of your right hand in the direction of the current, and your fingers will curl in the direction of the magnetic field.
Ampere's Circuital Law
Ampere's Circuital Law relates the integrated magnetic field around a closed loop to the electric current passing through the loop. Mathematically, it is expressed as:
$$ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}} $$
where $\oint \mathbf{B} \cdot d\mathbf{l}$ is the line integral of the magnetic field around a closed path, and $I_{\text{enc}}$ is the total current enclosed by the path.
Common MistakeA common mistake is to apply Ampere's Circuital Law to any arbitrary path. It is most effective for paths with high symmetry, such as circular or rectangular loops around straight or solenoid currents.
Magnetic Field Due to a Straight Conductor
For a long, straight conductor carrying a current $I$, the magnetic field at a distance $r$ from the conductor is given by:
$$ B = \frac{\mu_0 I}{2\pi r} $$
ExampleCalculate the magnetic field 2 cm away from a long, straight conductor carrying a current of 5 A. $$ B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7} \text{ T m/A})(5 \text{ A})}{2\pi (0.02 \text{ m})} = 5 \times 10^{-5} \text{ T} $$
Magnetic Field Due to a Circular Loop
The magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is given by:
$$ B = \frac{\mu_0 I}{2R} $$
ExampleFor a circular loop of radius 0.1 m carrying a current of 3 A, the magnetic field at the center is: $$ B = \frac{\mu_0 I}{2R} = \frac{(4\pi \times 10^{-7} \text{ T m/A})(3 \text{ A})}{2(0.1 \text{ m})} = 6 \times 10^{-6} \text{ T} $$
Magnetic Field Due to a Solenoid
A solenoid is a coil of wire with many turns, and the magnetic field inside a long solenoid carrying a current $I$ is uniform and given by:
$$ B = \mu_0 n I $$
where $n$ is the number of turns per unit length.
NoteThe magnetic field outside a long solenoid is nearly zero.
ExampleFor a solenoid with 1000 turns per meter carrying a current of 2 A, the magnetic field inside the solenoid is: $$ B = \mu_0 n I = (4\pi \times 10^{-7} \text{ T m/A})(1000 \text{ m}^{-1})(2 \text{ A}) = 2.5 \times 10^{-3} \text{ T} $$
Force Between Two Parallel Currents
Two parallel conductors carrying currents $I_1$ and $I_2$ separated by a distance $d$ exert a force on each other. The force per unit length is given by:
$$ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} $$
ExampleTwo parallel wires 1 cm apart carry currents of 10 A and 15 A. The force per unit length between them is: $$ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7} \text{ T m/A})(10 \text{ A})(15 \text{ A})}{2\pi (0.01 \text{ m})} = 3 \times 10^{-3} \text{ N/m} $$
TipRemember that parallel currents attract, while anti-parallel currents repel.
Conclusion
Understanding the relationship between moving charges and magnetism is fundamental in physics. By mastering the concepts and equations discussed, you will be well-prepared to tackle related problems in the NEET exam. Always remember to apply the right-hand rule for determining directions and to use symmetry when applying Ampere's Circuital Law.
