Theory, worked examples and 37 previous-year questions with full solutions — IIT JAM, GATE, CSIR NET-JRF and JEST, 2020 onwards.
Theory
Magnetic Forces: The Lorentz Force Law
The magnetic force on a charge \(Q\) moving with velocity \(\mathbf{v}\) in a
magnetic field \(\mathbf{B}\) is:
\[
\mathbf{F}_{\text{mag}} = Q(\mathbf{v}\times\mathbf{B}).
\]
In the presence of both electric and magnetic fields, the total force on \(Q\)
is the Lorentz force:
\[
\boxed{\mathbf{F} = Q\bigl[\mathbf{E} + (\mathbf{v}\times\mathbf{B})\bigr].}
\]
Key property:
Since \(\mathbf{F}_{\text{mag}}\perp\mathbf{v}\) always, the magnetic force
does no work on the charge. The kinetic energy and speed remain
constant; only the direction of motion changes.
Motion of a Charged Particle in a Uniform Magnetic Field
When a charged particle moves perpendicular to a uniform field
\(\mathbf{B}\), the magnetic force provides the centripetal acceleration,
so the particle follows a circular path (see figure):
\[
qvB = \frac{mv^2}{r}
\;\Rightarrow\;
\boxed{r = \frac{mv}{qB}.}
\]
The period of circular motion is:
\[
T = \frac{2\pi r}{v} = \frac{2\pi m}{qB},
\]
which is independent of the speed—a key property exploited in cyclotrons.
Circular motion of a negatively charged particle in a field
directed into the page. The magnetic force is always perpendicular
to the velocity.
When the velocity has a component parallel to \(\mathbf{B}\), that
component is unaffected (no force), while the perpendicular component drives
circular motion. The result is helical motion
(see figure):
\[
v_\perp = v\sin\theta, \qquad v_\| = v\cos\theta,
\]
where \(\theta\) is the angle between \(\mathbf{v}\) and \(\mathbf{B}\). The
pitch (axial advance per revolution) is:
\[
p = v_\| T = v\cos\theta\cdot\frac{2\pi m}{qB}.
\]
Helical motion: the perpendicular velocity component drives
circular motion; the parallel component translates the circle
along \(\mathbf{B}\). The pitch \(p\) is the advance per revolution.
Example 2.1. A proton enters a uniform field
\(B = 1.0\times10^{-4}\,\mathrm{T}\) at speed
\(v = 5\times10^5\,\mathrm{m/s}\). For what angle \(\theta\) between
\(\mathbf{v}\) and \(\mathbf{B}\) does the pitch equal the radius?
A current-carrying wire in a magnetic field experiences a force. For a wire
segment carrying current \(I\) in direction \(d\mathbf{l}\)
(see figure):
\[
d\mathbf{F} = I\,d\mathbf{l}\times\mathbf{B}.
\]
(a) Magnetic field lines around a current-carrying wire in the
plane of the page; field perpendicular to the page. (b) Circular
field lines around a long straight wire.
Derivation: In a wire of cross-section \(A\) with \(n\) charge carriers
per unit volume and drift velocity \(\mathbf{v}_d\), the current is
\(I = neAv_d\). The force on \(nA\,dl\) carriers in a segment \(dl\) is:
\[
d\mathbf{F} = (nA\,dl)\,e\,\mathbf{v}_d\times\mathbf{B}
= I\,d\mathbf{l}\times\mathbf{B}. \quad\checkmark
\]
An infinitesimal current-carrying wire element \(d\mathbf{l}\) in
a magnetic field \(\mathbf{B}\).
Example 2.2. A wire of length \(l = 50\,\mathrm{cm}\), mass
\(m = 10\,\mathrm{g}\), is suspended horizontally. A field
\(B = 0.50\,\mathrm{T}\) is applied. What current removes the tension?
Solution: Balance magnetic force against gravity:
\[
mg = IlB
\;\Rightarrow\;
I = \frac{mg}{lB}
= \frac{(0.010)(9.8)}{(0.50)(0.50)}
= 0.39\,\mathrm{A}.
\]
Example 2.3. A circular current loop (radius \(R\), current \(I\)) lies in
the \(xy\)-plane in a uniform field \(\mathbf{B} = B\hat{y}\). Find the force on
the upper half, lower half, and the total.
Solution: On a segment at angle \(\theta\),
\(dF = IBR\sin\theta\,d\theta\).
The net force on any closed current loop in a uniform magnetic field
is zero.
Force and Torque on a Current Loop
A closed current loop is a magnetic dipole with dipole moment:
\[
\mathbf{m} = NIA\hat{\mathbf{n}},
\]
where \(\hat{\mathbf{n}}\) is the right-hand normal to the loop, \(N\) is the
number of turns, and \(A\) is the area. The torque in a uniform field is:
\[
\boldsymbol{\tau} = \mathbf{m}\times\mathbf{B},
\]
and the potential energy is:
\[
U = -\mathbf{m}\cdot\mathbf{B}.
\]
Example 2.4. A circular loop (\(r = 2.0\,\mathrm{cm}\),
\(I = 2.0\,\mathrm{mA}\)) makes angle \(30\ensuremath{^\circ}\) with a field
\(B = 0.50\,\mathrm{T}\).
(a) Find \(|\mathbf{m}|\). (b) Find \(\tau\) and \(U\).
Solution:
(a) \(m = IA = (2.0\times10^{-3})(\pi\times0.02^2) = 2.5\times10^{-6}\,\mathrm{A\cdot m^2}\).
Example 2.5 (Cyclotron). Alpha particles
(\(m = 6.64\times10^{-27}\,\mathrm{kg}\),
\(q = 3.2\times10^{-19}\,\mathrm{C}\)) are accelerated in a cyclotron of
radius \(R = 0.50\,\mathrm{m}\), field \(B = 1.8\,\mathrm{T}\).
(a) Find the period. (b) Find the maximum kinetic energy.
(b) At maximum radius,
\(v_{\max} = \dfrac{qBR}{m}\), so:
\[
K_{\max} = \frac{1}{2}mv_{\max}^2
= \frac{q^2B^2R^2}{2m}
= \frac{(3.2\times10^{-19})^2(1.8)^2(0.50)^2}{2(6.64\times10^{-27})}
= 6.2\times10^{-12}\,\mathrm{J} = 39\,\mathrm{MeV}.
\]
The Biot-Savart Law
The magnetic field at a field point \(P\) due to a steady current \(I\) in a
wire element \(d\mathbf{l}\) is given by the Biot-Savart law
(see figure):
\[
\boxed{d\mathbf{B} = \frac{\mu_0}{4\pi}\frac{I\,d\mathbf{l}\times\hat{\mathbf{r}}}{r^2},}
\]
where \(\hat{\mathbf{r}}\) points from the source element \(d\mathbf{l}\) to the
field point \(P\), and
\(\mu_0 = 4\pi\times10^{-7}\,\mathrm{T\cdot m/A}\) is the permeability of
free space.
Biot-Savart law: current element \(I\,d\mathbf{l}\) produces field
\(d\mathbf{B}\) at point \(P\). The vector \(\hat{\mathbf{r}}\) points
from the source to the field point.
Example 2.6. A short wire segment (\(\Delta l = 1.0\,\mathrm{cm}\),
\(I = 2.0\,\mathrm{A}\), vertical) is located \(1\,\mathrm{m}\) from point \(P\)
in the horizontal direction (see figure). Find \(B\) at \(P\).
A short current element; the field point is at horizontal
distance \(x = 1\,\mathrm{m}\).
Solution: Since \(\Delta l \ll x\), we can treat this as a point source.
The angle between \(\Delta\mathbf{l}\) and \(\hat{\mathbf{r}}\) is:
\[
\theta = \tan^{-1}\!\left(\frac{x}{\Delta l}\right)
= \tan^{-1}\!\left(\frac{1\,\mathrm{m}}{0.01\,\mathrm{m}}\right)
= 89.4\ensuremath{^\circ}.
\]
\[
B = \frac{\mu_0}{4\pi}\frac{I\,\Delta l\sin\theta}{r^2}
= (10^{-7})\frac{(2)(0.01)\sin89.4\ensuremath{^\circ}}{(1)^2}
= 2.0\times10^{-9}\,\mathrm{T},
\quad\text{directed into the page.}
\]
Example 2.7. Find the magnetic field at the centre of a circular arc
of radius \(R\) subtending angle \(\theta_0\) and carrying current \(I\)
(see figure).
A circular arc of radius \(R\) subtending angle \(\theta_0\)
(in radians).
Solution: At the centre, \(d\mathbf{l}\perp\hat{\mathbf{r}}\) always,
so \(|d\mathbf{l}\times\hat{\mathbf{r}}| = dl = R\,d\theta\):
\[
B = \frac{\mu_0 I}{4\pi R}\int_0^{\theta_0}d\theta
= \frac{\mu_0 I\,\theta_0}{4\pi R}.
\]
For a full circle (\(\theta_0 = 2\pi\)): \(B = \mu_0 I/(2R)\).
Example 2.8 (Finite straight wire). Find the magnetic field at
perpendicular distance \(R\) from a finite straight wire carrying current \(I\),
for a wire segment subtending angles \(\theta_1\) and \(\theta_2\) from the
foot of the perpendicular (see figure).
Finite straight wire; \(\theta_1\) and \(\theta_2\) are the angles to
the ends of the wire from the perpendicular foot.
Solution: With \(r = \sqrt{x^2+R^2}\) and
\(\sin\alpha = R/\sqrt{x^2+R^2}\):
\[
B = \frac{\mu_0 I}{4\pi}\int_{\text{wire}}\frac{\sin\alpha\,dx}{r^2}
= \frac{\mu_0 I}{2\pi R}\int_0^\infty\frac{R\,dx}{(x^2+R^2)^{3/2}}.
\]
For a general finite wire subtending angles \(\theta_1\) (below
perpendicular foot) and \(\theta_2\) (above):
\[
\boxed{B = \frac{\mu_0 I}{4\pi R}(\sin\theta_2 - \sin\theta_1).}
\]
For an infinite wire (\(\theta_1 \to -90\ensuremath{^\circ}\), \(\theta_2\to 90\ensuremath{^\circ}\)):
\[
B = \frac{\mu_0 I}{4\pi R}(1+1) = \frac{\mu_0 I}{2\pi R}.
\]
Right-hand rule: curl the fingers in the direction of current;
the thumb points along \(\mathbf{B}\).
Example 2.9. Three wires at corners of a square (side
\(a = 0.01\,\mathrm{m}\)), all carrying \(I = 2\,\mathrm{A}\) into the page
(see figure). Find \(\mathbf{B}\) at the fourth corner \(P\).
Three wires at corners of a square; \(B\) is found at the fourth
corner \(P\).
Solution:
Vector diagram of the three magnetic field contributions at \(P\).
Wires 1 and 3 are at distance \(a\); wire 2 is at diagonal distance
\(a\sqrt{2}\).
(a) Field of wire 1 at wire 2. (b) Force between two parallel
wires is attractive for parallel currents, repulsive for
anti-parallel currents.
Wire 1 produces field \(B_1 = \mu_0 I_1/2\pi r\) at wire 2. The force per
unit length on wire 2 is:
\[
\frac{F}{l} = I_2 B_1 = \frac{\mu_0 I_1 I_2}{2\pi r}.
\]
Force is attractive for parallel currents, repulsive for
anti-parallel.
Magnetic Field of a Current Loop
Field at point \(P\) on the axis of a circular current loop of
radius \(R\) at axial distance \(y\).
For a loop of radius \(R\) in the \(xz\)-plane carrying current \(I\), the field
at axial distance \(y\) (see figure):
By symmetry, transverse components cancel. The axial component from each
element \(dl\) has \(\cos\theta = R/\sqrt{y^2+R^2}\):
\[
\mathbf{B} = \frac{\mu_0 IR^2}{2(y^2+R^2)^{3/2}}\,\hat{\mathbf{y}}.
\]
In terms of the magnetic moment \(\mathbf{m} = I\pi R^2\hat{\mathbf{y}}\):
\[
\mathbf{B} = \frac{\mu_0}{2\pi}\frac{\mathbf{m}}{(y^2+R^2)^{3/2}}.
\]
At the centre (\(y = 0\)):
\[
\mathbf{B} = \frac{\mu_0 I}{2R}\,\hat{\mathbf{y}}.
\]
For a flat coil of \(N\) turns:
\[
B = \frac{\mu_0 NI}{2R}.
\]
In the far-field limit (\(y \gg R\)), the field reduces to the magnetic dipole
form:
\[
\mathbf{B} \approx \frac{\mu_0}{2\pi}\frac{\mathbf{m}}{y^3}.
\]
Note:
Energy of a magnetic dipole: \(U = -\mathbf{m}\cdot\mathbf{B}\).
Interaction energy of two magnetic dipoles at separation \(\mathbf{r}\):
\[
U = \frac{\mu_0}{4\pi r^3}
\bigl[\mathbf{m}_1\cdot\mathbf{m}_2
- 3(\mathbf{m}_1\cdot\hat{\mathbf{r}})
(\mathbf{m}_2\cdot\hat{\mathbf{r}})\bigr].
\]
Ampère's Law
For steady currents, Ampère's law states:
\[
\boxed{\oint_C \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_{\text{enc}},}
\]
where \(I_{\text{enc}}\) is the total steady current threading any open surface
bounded by the closed path \(C\). Only currents inside the path
contribute.
In differential form (from Stokes's theorem):
\[
\nabla\times\mathbf{B} = \mu_0\mathbf{J}.
\]
Problem-Solving Strategy:
Identify the symmetry (spherical, cylindrical, planar).
If none, use Biot-Savart.
Use the right-hand rule to determine the direction of \(\mathbf{B}\).
Choose an Amperian loop where \(\mathbf{B}\) is constant and parallel
(or perpendicular and zero) everywhere on the loop.
Evaluate \(I_{\text{enc}}\) inside the loop.
Compute \(\oint\mathbf{B}\cdot d\mathbf{l}\) and equate to
\(\mu_0 I_{\text{enc}}\).
Example 2.10 — Infinite straight wire.
For a circular path of radius \(r\) centred on the wire:
\[
\oint\mathbf{B}\cdot d\mathbf{l} = B_\theta(2\pi r) = \mu_0 I
\;\Rightarrow\;
\boxed{B = \frac{\mu_0 I}{2\pi r}.}
\]
Example 2.11 — Infinite solenoid. (\(n\) turns/m, current \(I\)):
Use a rectangular Amperian loop with one side inside and one outside.
The field outside is zero; the field inside is uniform:
\[
Bl = \mu_0\,nl\,I
\;\Rightarrow\;
\boxed{B = \mu_0 nI \quad \text{(inside)}, \qquad B = 0 \quad\text{(outside)}.}
\]
Example 2.12 — Toroid. (N total turns, inner radius \(a\), outer
radius \(b\)):
\[
B = \begin{cases}
\dfrac{\mu_0 NI}{2\pi r} & a < r < b, \\[4pt]
0 & r < a \text{ or } r > b.
\end{cases}
\]
Example 2.13 — Thick wire. (radius \(a\), uniform current density,
total current \(I_0\)):
Cross-section of a thick wire of radius \(a\) carrying current
\(I_0\); Amperian loop shown at radius \(r\).\(B\) vs \(r\) for a thick wire: linear inside, \(1/r\) outside.
For \(r\leq a\): enclosed current \(= (r^2/a^2)I_0\):
\[
B(2\pi r) = \mu_0\frac{r^2}{a^2}I_0
\;\Rightarrow\;
B = \frac{\mu_0 I_0\,r}{2\pi a^2} \quad (r\leq a).
\]
For \(r\geq a\):
\[
B = \frac{\mu_0 I_0}{2\pi r} \quad (r\geq a).
\]
The field grows linearly inside and falls as \(1/r\) outside (see figure).
Concept Checks — Magnetic Vector Potential & Dipole
Concept check.
Magnetic dipole moment of disc (radius \(R\), surface charge \(\sigma\)) rotating at \(\omega\):
Show answer
Consider ring at \(r\): \(dI=\sigma\omega r\,dr\); \(dm=\pi r^2 dI=\pi r^2\sigma\omega r\,dr\). Integrate: \(m=\pi\sigma\omega\int_0^R r^3\,dr=\pi\sigma\omega R^4/4\).
Magnetism in Matter
On a microscopic level, every electron in an atom contributes a magnetic
dipole moment through two mechanisms: its orbital motion around the nucleus,
and its intrinsic spin. The net atomic magnetic dipole moment is the vector
sum of all these contributions.
Atom
Magnetic Moment (\(10^{-24}\,\mathrm{A\cdot m^2}\))
H
9.27
He
0
Li
9.27
O
13.9
Na
9.27
S
13.9
Without an external field, thermal agitation randomizes the orientations of
atomic dipoles, giving zero net magnetization. An applied field tends to
align them. The competition between field alignment and thermal disorder
determines the magnetic response.
Classification of Magnetic Materials
Paramagnetic: Atoms have permanent dipole moments. Applied field
causes partial alignment (enhanced by field). The additional field
\(\mathbf{B}_m\) is parallel to the applied field \(\mathbf{B}_0\), so the
total field is enhanced.
Diamagnetic: Atoms have no permanent moments, but the applied
field induces moments opposite to itself (Lenz's law at the atomic
level). Net effect is a slight weakening of the applied field.
\(\chi < 0\).
Ferromagnetic: Quantum exchange interactions lock large numbers of
dipoles (magnetic domains) into parallel alignment. Very large
\(\chi\) (order \(10^3\)–\(10^4\)); permanent magnetization possible.
Magnetic Susceptibility and Permeability
For paramagnetic and diamagnetic materials in a solenoid, the total field is:
\[
\mathbf{B} = \mathbf{B}_0 + \mathbf{B}_m = (1+\chi)\mathbf{B}_0,
\]
where \(\chi\) is the magnetic susceptibility. The
magnetic permeability is:
\[
\mu_m = (1+\chi)\mu_0,
\qquad\text{so}\qquad
B = \mu_m nI
\]
for a filled solenoid. (\(\mu_m\) should not be confused with the magnetic
dipole moment \(\mathbf{m}\).)
Paramagnetic
\(\chi\)
Diamagnetic
\(\chi\)
Aluminum
\(2.2\times10^{-5}\)
Bismuth
\(-1.7\times10^{-5}\)
Calcium
\(1.4\times10^{-5}\)
Carbon (diamond)
\(-2.2\times10^{-5}\)
Chromium
\(3.1\times10^{-4}\)
Copper
\(-9.7\times10^{-6}\)
Magnesium
\(1.2\times10^{-5}\)
Lead
\(-1.8\times10^{-5}\)
O\(_2\) (1 atm)
\(1.8\times10^{-6}\)
Mercury
\(-2.8\times10^{-5}\)
O\(_2\) liq (90 K)
\(3.5\times10^{-3}\)
H\(_2\) (1 atm)
\(-2.2\times10^{-9}\)
Tungsten
\(6.8\times10^{-5}\)
N\(_2\) (1 atm)
\(-6.7\times10^{-9}\)
Air (1 atm)
\(3.6\times10^{-7}\)
Water
\(-9.1\times10^{-6}\)
Paramagnetic Alignment
Alignment of magnetic dipoles in a paramagnetic material inside
a solenoid. (a) No field: random orientation. (b) With applied
field: partial alignment. (c) Equivalent representation. (d)
Effective surface current producing the enhanced field.
The energy difference between a dipole aligned with and against a field \(B\)
is \(U_B = 2mB\). For \(m \approx 9.3\times10^{-24}\,\mathrm{A\cdot m^2}\) and
\(B = 1\,\mathrm{T}\):
\[
U_B = 1.9\times10^{-23}\,\mathrm{J},
\]
compared to thermal energy at \(27\ensuremath{^\circ}C\):
\[
k_BT \approx 4.1\times10^{-21}\,\mathrm{J} \approx 220\,U_B.
\]
Thermal energy greatly exceeds the alignment energy, so only a very small
fraction of dipoles are aligned at room temperature—explaining why \(\chi\)
is small for paramagnets.
Ferromagnetism and Hysteresis
(a) Random domain orientation (unmagnetized). (b) Partial domain
alignment in an applied field. (c) Domain structure in a nickel
crystal.
Ferromagnetic materials contain magnetic domains—microscopic
regions (\(10^{-12}\) to \(10^{-8}\,\mathrm{m}^3\)) within which quantum
exchange coupling aligns all dipoles. The net field can be thousands of
times \(B_0\).
Example 2.14 (Iron core). A solenoid with \(n = 20\,\mathrm{turns/cm}\)
(\(= 2000\,\mathrm{m}^{-1}\)) and current \(I_0 = 0.20\,\mathrm{A}\) is wound
around an iron cylinder (see figure).
Magnetization curve (\(B\) vs \(B_0\)) for annealed iron.
Hysteresis loop for a ferromagnet. Path \(U\to a\): initial
magnetization. \(a\to b\): reversed applied field.
Remanence: \(B \neq 0\) when \(B_0 = 0\) (permanent magnet).
Coercivity: the reverse \(B_0\) needed to demagnetize.Comparison of magnetic field lines for a finite solenoid (left)
and a bar magnet (right)—the similarity confirms the domain
model.
When \(B_0\) is cycled, the iron follows the hysteresis loop of
the figure. Key features:
Remanence: \(B \neq 0\) when \(B_0 = 0\) (permanent magnet).
Coercivity: the reverse applied field needed to reduce
\(B\) back to zero.
The area enclosed by the loop equals the energy dissipated per cycle.
A ferromagnetic material at temperature above Curie temperature behaves as:
(a) Ferromagnet (b) Diamagnet (c) Paramagnet (d) Antiferromagnet
Show answer
(c). Above the Curie temperature, ferromagnetic order breaks down and material becomes paramagnetic (Curie-Weiss law).
Faraday's Law
The magnetic flux through a surface \(S\) is:
\[
\Phi_m = \int_S \mathbf{B}\cdot\hat{\mathbf{n}}\,dA.
\]
Magnetic flux: the integral of \(\mathbf{B}\) over a surface.
Flux is maximum when \(\mathbf{B}\parallel\hat{\mathbf{n}}\).
Faraday's law states that the emf induced in a loop is:
\[
\boxed{\mathcal{E} = -\frac{d\Phi_m}{dt}.}
\]
For a coil of \(N\) turns:
\[
\mathcal{E} = -N\frac{d\Phi_m}{dt}.
\]
Flux can change through three mechanisms: (1) changing \(|\mathbf{B}|\),
(2) changing the area, (3) changing the orientation. The negative sign
reflects Lenz's law (next section).
Example 2.15. A square coil (\(l = 0.25\,\mathrm{m}\), \(N = 200\),
\(R = 5.0\,\Omega\)) is in a field decreasing at
\(dB/dt = -0.040\,\mathrm{T/s}\).
(a) Find \(|\mathcal{E}|\). (b) Find \(I\).
Solution:
(a) \(|\mathcal{E}| = N l^2 |dB/dt| = (200)(0.25)^2(0.040) = 0.50\,\mathrm{V}\).
Lenz's law: The direction of the induced emf (and hence the induced
current) is such as to oppose the change in flux that caused it.
(a) Approaching north pole: induced current opposes increasing
flux (counterclockwise as viewed from the magnet).
(b) Approaching south pole: induced current clockwise.
Lenz's law is a consequence of energy conservation: if the induced current
enhanced the flux change instead of opposing it, the system would
spontaneously accelerate, violating energy conservation.
Example 2.16.
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