Chapter 4: Moving Charges & Magnetism
1. Oersted's Experiment (The Origin)
In 1820, Hans Christian Oersted discovered the link between electricity and magnetism. He observed that a compass needle placed near a current-carrying wire experienced deflection.
Moving charges (Current) produce a Magnetic Field ($\vec{B}$) in the surrounding space.
Stationary charges produce only Electric Field ($\vec{E}$).
2. Finding the Direction (Right Hand Rule)
The direction of the magnetic field lines around a straight conductor is given by the Right Hand Thumb Rule.
[attachment_0](attachment)The Rule: Imagine holding the current-carrying conductor in your right hand such that the Thumb points in the direction of the Current ($I$).
Then, the Curled Fingers will point in the direction of the Magnetic Field Lines ($\vec{B}$).
Notation for Field Direction:
$\odot$ (Dot): Magnetic field coming OUT of the page.
$\otimes$ (Cross): Magnetic field going INTO the page.
3. Magnetic Field due to Straight Wire
Consider a straight wire carrying current $I$. We wish to find the magnetic field $B$ at a point $P$ at a perpendicular distance $r$.
Case A: Infinitely Long Wire
For a wire of infinite length, the magnetic field is given by:
($\mu_0 = 4\pi \times 10^{-7}$ T$\cdot$m/A is Permeability of free space)
Case B: Finite Length Wire
If the wire subtends angles $\phi_1$ and $\phi_2$ at the observation point $P$:
Note: For infinite wire, $\phi_1 = \phi_2 = 90^\circ$. Since $\sin 90^\circ = 1$, the formula reduces to Case A.
Chapter 4: Moving Charges & Magnetism
1. Magnetic Field at the Center of Loop
Consider a circular coil of radius $R$ carrying current $I$. The field at the geometric center $O$ is perpendicular to the plane of the coil.
For a coil with $N$ turns:
$$B = \frac{\mu_0 N I}{2R}$$2. Magnetic Field on the Axis (Derivation)
This is a 5-Mark Derivation in Board Exams.
Consider a point $P$ on the axis of the loop at a distance $x$ from the center $O$. The radius of the loop is $R$.
Formula:
Special Cases:
1. At Center ($x=0$):
$$B = \frac{\mu_0 I R^2}{2(R^2)^{3/2}} = \frac{\mu_0 I R^2}{2R^3} = \frac{\mu_0 I}{2R}$$
(Matches the center formula ✅)
2. At Large Distance ($x \gg R$):
$$B \approx \frac{\mu_0 I R^2}{2x^3} = \frac{\mu_0}{4\pi} \frac{2M}{x^3}$$
(where $M = I \cdot \pi R^2$ is Magnetic Dipole Moment).
This shows a current loop behaves like a Magnetic Dipole.
3. Variation Graph (B vs x)
The magnetic field is maximum at the center and decreases as we move away along the axis.
Chapter 4: Moving Charges & Magnetism
1. The Solenoid
Definition: A solenoid consists of a long wire wound in the form of a helix where the neighboring turns are closely spaced. When current flows through it, it behaves like a bar magnet.
Properties of Magnetic Field:
1. Inside: The field is uniform, strong, and directed along the axis.
2. Outside: The field is extremely weak (approaches zero for an ideal solenoid).
Formula for Magnetic Field
For a long ideal solenoid carrying current $I$:
Where $n$ is the number of turns per unit length.
$$n = \frac{N}{L}$$
Note: At the ends of a finite solenoid, the magnetic field is half of that at the center: $B_{end} = \frac{\mu_0 n I}{2}$.
2. The Toroid
Definition: A toroid is essentially a hollow circular ring on which a large number of turns of a wire are closely wound. It can be viewed as a solenoid which has been bent into a circle.
Magnetic Field:
1. Inside the core: Uniform magnetic field exists.
2. In the open space (interior/exterior): Field is Zero.
$$B = \mu_0 n I$$
Where $n = \frac{N}{2\pi r}$ ($r$ is the average radius of the toroid).
Chapter 4: Question Bank
Section A: MCQs (1 Mark)
Q1. A current carrying power line carries current from West to East. What is the direction of the magnetic field 1m above it?
(a) North to South (b) South to North (c) East to West (d) West to East
Ans: (a) North to South (Using Right Hand Thumb Rule).
Q2. The magnetic field at the center of a circular coil of radius r is B. If the current and radius both are doubled, the new magnetic field will be:
(a) 2B (b) B (c) B/2 (d) 4B
Ans: (b) B
Reason: $B = \frac{\mu_0 I}{2R}$. If $I \to 2I, R \to 2R$, then $B' = \frac{\mu_0 (2I)}{2(2R)} = B$.
Section B: Important Numericals (3 Marks)
Q3. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Solution:
Given: $I = 35$ A, $r = 20$ cm = $0.2$ m.
Formula: $B = \frac{\mu_0 I}{2\pi r}$
Since $\frac{\mu_0}{4\pi} = 10^{-7}$, then $\frac{\mu_0}{2\pi} = 2 \times 10^{-7}$.
$B = \frac{2 \times 10^{-7} \times 35}{0.2} = \frac{70 \times 10^{-7}}{0.2}$
$B = 350 \times 10^{-7} = 3.5 \times 10^{-5}$ T.
Ans: $3.5 \times 10^{-5}$ Tesla.
Q4. A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnetic field inside the solenoid?
Solution:
Given: $N = 500$, $L = 0.5$ m, $I = 5$ A.
Calculate $n$ (turns per unit length): $n = \frac{N}{L} = \frac{500}{0.5} = 1000$ turns/m.
Formula: $B = \mu_0 n I$
$B = 4\pi \times 10^{-7} \times 1000 \times 5$
$B = 20\pi \times 10^{-4} = 62.8 \times 10^{-4}$ T.
Ans: $6.28 \times 10^{-3}$ Tesla.
Section C: Conceptual (2 Marks)
Q5. Why does the magnetic field inside a toroid not depend on the radius of the cross-section?
Ans: Because the magnetic field inside an ideal toroid depends on the turn density ($n$) and current ($I$) ($B = \mu_0 n I$). It is uniform throughout the core for a large radius toroid.
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