Chapter Notes

Introduction

For over 2000 years, electricity and magnetism were seen as separate phenomena. It wasn't until 1820 that their deep connection was discovered. Danish physicist Hans Christian Oersted, during a lecture, noticed that an electric current flowing through a wire deflected a nearby magnetic compass needle.

This simple observation was groundbreaking. Oersted found that:

• The compass needle aligns itself tangentially to a circle with the current-carrying wire at its center.
• Reversing the current's direction reverses the needle's orientation.
• The deflection is stronger if the current is larger or if the needle is closer to the wire.
• Iron filings sprinkled around the wire form concentric circles.

Oersted's conclusion was revolutionary: Moving charges (or electric currents) produce a magnetic field in the space around them. This discovery paved the way for James Maxwell to unify the laws of electricity and magnetism in 1864, leading to the understanding of light as an electromagnetic wave and the technological advancements of the 20th century.

• A dot (⊙) represents a current or field coming out of the page, like the tip of an arrow coming towards you.
• A cross (⊗) represents a current or field going into the page, like the feathered tail of an arrow moving away from you.

Magnetic Force

Sources and Fields

Just as a static electric charge creates an electric field E in the space around it, a moving charge or a current creates a magnetic field B.

• An electric field is produced by a source charge, $Q$. Another charge, $q$, placed in this field experiences an electric force, $\mathbf{F} = q\mathbf{E}$.
• Similarly, a magnetic field is produced by a source current (moving charges). Another moving charge placed in this field will experience a magnetic force.

Both electric and magnetic fields obey the principle of superposition, which means the total field at any point is the vector sum of the fields from all individual sources.

Magnetic Field, Lorentz Force

When a charge $q$ moves with velocity v through a region with both an electric field E and a magnetic field B, the total force it experiences is called the Lorentz force.

The Lorentz force is given by the equation: $\mathbf{F} = q[\mathbf{E}(\mathbf{r}) + \mathbf{v} \times \mathbf{B}(\mathbf{r})]$

This total force is the sum of the electric force ($\mathbf{F}_{\text{electric}} = q\mathbf{E}$) and the magnetic force ($\mathbf{F}_{\text{magnetic}} = q(\mathbf{v} \times \mathbf{B})$).

Let's look at the key features of the magnetic force:

• It depends on the charge ($q$), velocity ($\mathbf{v}$), and magnetic field ($\mathbf{B}$). A negative charge experiences a force in the opposite direction to a positive charge.
• The force is a result of a vector product ($\mathbf{v} \times \mathbf{B}$). This means the magnetic force is always perpendicular to both the velocity of the charge and the magnetic field. Its direction can be found using the right-hand rule.
• The force is zero if the charge is stationary ($v=0$). Only moving charges experience a magnetic force.
• The force is zero if the velocity is parallel or anti-parallel to the magnetic field, because the vector product of parallel vectors is zero.

The SI unit of magnetic field is the tesla (T). One tesla is the magnetic field in which a charge of 1 Coulomb, moving at 1 m/s perpendicular to the field, experiences a force of 1 Newton. The tesla is a large unit; a smaller, non-SI unit called the gauss is also used ($1 \text{ gauss} = 10^{-4} \text{ T}$). The Earth's magnetic field is about $3.6 \times 10^{-5} \text{ T}$.

Magnetic Force on a Current-Carrying Conductor

We can extend the concept of magnetic force from a single moving charge to a conductor carrying a current. A current is simply a collection of many moving charges.

Consider a straight rod of length $l$ and cross-sectional area $A$, carrying a current $I$. If this rod is placed in an external uniform magnetic field B, it experiences a force.

The force on a current-carrying conductor is given by: $\mathbf{F} = I\boldsymbol{l} \times \mathbf{B}$ Here, $\boldsymbol{l}$ is a vector whose magnitude is the length of the rod, $l$, and whose direction is the direction of the current $I$.

If the wire is not straight, we can imagine it as a collection of tiny straight segments $d\boldsymbol{l}$ and sum (or integrate) the forces on each segment to find the total force.

Given

• Mass, $m = 200 \text{ g} = 0.2 \text{ kg}$
• Length, $l = 1.5 \text{ m}$
• Current, $I = 2 \text{ A}$
• Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find

The magnitude of the magnetic field, $B$.

Formula

For the wire to be suspended in mid-air, the upward magnetic force must balance the downward gravitational force (weight). $F_{\text{magnetic}} = F_{\text{gravity}}$ $IlB = mg$

Solution

Rearranging the formula to solve for B: $B = \frac{mg}{Il}$ Substitute the given values into the formula: $B = \frac{0.2 \text{ kg} \times 9.8 \text{ m/s}^2}{2 \text{ A} \times 1.5 \text{ m}} = \frac{1.96}{3.0} \text{ T} \approx 0.65 \text{ T}$

Final Answer The magnitude of the magnetic field required is $0.65 \text{ T}$.

Given

• Velocity $\mathbf{v}$ is along the positive x-axis.
• Magnetic field $\mathbf{B}$ is along the positive y-axis.

To Find

The direction of the Lorentz force $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$ for (a) an electron and (b) a proton.

Formula

The direction is determined by the right-hand rule for the vector product $\mathbf{v} \times \mathbf{B}$, and then adjusted for the sign of the charge $q$.

Solution

Using the right-hand rule, if you point your fingers along the x-axis ($\mathbf{v}$) and curl them towards the y-axis ($\mathbf{B}$), your thumb points along the positive z-axis. So, the direction of $\mathbf{v} \times \mathbf{B}$ is along the +z axis.

(a) For an electron The charge $q$ is negative. Therefore, the force $\mathbf{F}$ is in the opposite direction to $\mathbf{v} \times \mathbf{B}$.

Answer for part (a) = The force is along the -z axis.

(b) For a proton The charge $q$ is positive. Therefore, the force $\mathbf{F}$ is in the same direction as $\mathbf{v} \times \mathbf{B}$.

Answer for part (b) = The force is along the +z axis.

Motion in a Magnetic Field

A key difference between electric and magnetic forces is their effect on a particle's energy.

• An electric force can do work on a charged particle, changing its kinetic energy (speed).
• A magnetic force does no work. This is because the force is always perpendicular to the particle's velocity. Since work is done only when a force has a component along the direction of motion, the magnetic force can change the direction of the particle's momentum, but not its magnitude (speed or kinetic energy).

Let's consider a charged particle moving in a uniform magnetic field B.

Case 1: Velocity is Perpendicular to the Magnetic Field

If a particle's velocity v is perpendicular to B, the magnetic force acts as a centripetal force, pulling the particle into a circular path.

• The magnetic force has a magnitude of $F = qvB$.
• The centripetal force required for circular motion of radius $r$ is $F_c = mv^2/r$.

Equating these two forces gives us the radius of the circular path: $qvB = \frac{mv^2}{r}$ $r = \frac{mv}{qB}$ This shows that the radius of the circle is proportional to the particle's momentum ($mv$).

The angular frequency ($\omega$) of this circular motion is: $\omega = \frac{v}{r} = \frac{qB}{m}$ The frequency of rotation, $\nu = \omega / 2\pi$, is called the cyclotron frequency.

Case 2: Velocity has a Component Parallel to the Magnetic Field

If the particle's velocity is not perpendicular to B, we can split the velocity vector into two components:

1. A component perpendicular to B ($v_{\perp}$): This causes circular motion, as described above.
2. A component parallel to B ($v_{||}$): This component is unaffected by the magnetic field, so the particle continues to move along the field direction with constant velocity $v_{||}$.

The combination of circular motion in one plane and straight-line motion perpendicular to that plane results in a helical path (like a corkscrew).

The distance the particle travels along the magnetic field direction in one complete revolution is called the pitch of the helix. $p = v_{||} T = v_{||} \frac{2\pi m}{qB}$ where $T$ is the time period of one revolution.

Given

• Mass of electron, $m = 9 \times 10^{-31} \text{ kg}$
• Charge of electron, $q = 1.6 \times 10^{-19} \text{ C}$
• Speed, $v = 3 \times 10^7 \text{ m/s}$
• Magnetic field, $B = 6 \times 10^{-4} \text{ T}$

To Find

(a) Radius of the path, $r$ (b) Frequency, $\nu$ (c) Energy in keV

Formula

$r = \frac{mv}{qB}$ $\nu = \frac{qB}{2\pi m}$ $E = \frac{1}{2}mv^2$

Solution

(a) Calculate the radius $r = \frac{(9 \times 10^{-31} \text{ kg}) \times (3 \times 10^7 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C}) \times (6 \times 10^{-4} \text{ T})} = \frac{27 \times 10^{-24}}{9.6 \times 10^{-23}} \text{ m}$ $r \approx 0.28 \text{ m} = 28 \text{ cm}$

Answer for part (a) = $28 \text{ cm}$

(b) Calculate the frequency $\nu = \frac{qB}{2\pi m} = \frac{(1.6 \times 10^{-19} \text{ C}) \times (6 \times 10^{-4} \text{ T})}{2\pi \times (9 \times 10^{-31} \text{ kg})} = \frac{9.6 \times 10^{-23}}{56.5 \times 10^{-31}} \text{ Hz}$ $\nu \approx 0.17 \times 10^8 \text{ Hz} = 17 \times 10^6 \text{ Hz} = 17 \text{ MHz}$

Answer for part (b) = $17 \text{ MHz}$

(c) Calculate the energy First, find the energy in Joules: $E = \frac{1}{2}mv^2 = \frac{1}{2} \times (9 \times 10^{-31} \text{ kg}) \times (3 \times 10^7 \text{ m/s})^2$ $E = 4.5 \times 10^{-31} \times 9 \times 10^{14} \text{ J} = 40.5 \times 10^{-17} \text{ J} \approx 4 \times 10^{-16} \text{ J}$ Now, convert to keV: $E = \frac{4 \times 10^{-16} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 2.5 \times 10^3 \text{ eV} = 2.5 \text{ keV}$

Answer for part (c) = $2.5 \text{ keV}$

Magnetic Field due to A Current Element, Biot-Savart Law

The Biot-Savart law is a fundamental law that describes the magnetic field generated by a steady electric current. It's the magnetic equivalent of Coulomb's law in electrostatics.

The law states that the magnetic field $d\mathbf{B}$ from an infinitesimal current element $I d\boldsymbol{l}$ at a distance $r$ is given by: $d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\boldsymbol{l} \times \mathbf{r}}{r^3}$

• $I$ is the current.
• $d\boldsymbol{l}$ is a vector representing a tiny segment of the wire, pointing in the direction of the current.
• $\mathbf{r}$ is the displacement vector from the current element to the point where the field is being calculated.
• $\mu_0$ is the permeability of free space, a fundamental constant with the value $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.

The magnitude of this field is: $|d\mathbf{B}| = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$ where $\theta$ is the angle between the current element $d\boldsymbol{l}$ and the displacement vector $\mathbf{r}$.

Comparing Biot-Savart Law and Coulomb's Law

• Similarities: Both are long-range inverse-square laws, and both obey the principle of superposition.
• Differences:
  • Source: The electric field source (charge) is a scalar, while the magnetic field source (current element $I d\boldsymbol{l}$) is a vector.
  • Direction: The electric field points along the line connecting the source and the point. The magnetic field is perpendicular to the plane containing the source element and the point.
  • Angle Dependence: The magnetic field's magnitude depends on the angle $\theta$, while the electric field from a point charge is spherically symmetric. The magnetic field is zero along the line of the current element itself ($\theta=0$).

Given

• Current element, $\Delta\boldsymbol{l} = \Delta x \hat{\mathbf{i}}$, with $\Delta x = 1 \text{ cm} = 10^{-2} \text{ m}$
• Current, $I = 10 \text{ A}$
• Position vector, $\mathbf{r} = 0.5 \hat{\mathbf{j}}$ m, so $r = 0.5 \text{ m}$
• The angle $\theta$ between $\Delta\boldsymbol{l}$ (along x-axis) and $\mathbf{r}$ (along y-axis) is $90^{\circ}$.

To Find

The magnetic field $d\mathbf{B}$ at the point $(0, 0.5, 0)$.

Formula

$|d\mathbf{B}| = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$ The direction is given by the cross product $\Delta\boldsymbol{l} \times \mathbf{r}$.

Solution

Calculate the magnitude: $|d\mathbf{B}| = (10^{-7} \text{ T m/A}) \frac{(10 \text{ A}) \times (10^{-2} \text{ m}) \times \sin(90^\circ)}{(0.5 \text{ m})^2}$ $|d\mathbf{B}| = \frac{10^{-7} \times 10 \times 10^{-2}}{0.25} \text{ T} = \frac{10^{-8}}{0.25} \text{ T} = 4 \times 10^{-8} \text{ T}$ Determine the direction: The cross product is $\Delta\boldsymbol{l} \times \mathbf{r} = (\Delta x \hat{\mathbf{i}}) \times (y \hat{\mathbf{j}}) = (\Delta x y) (\hat{\mathbf{i}} \times \hat{\mathbf{j}})$. Since $\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}$, the direction of the field is along the positive z-axis.

Final Answer The magnetic field is $4 \times 10^{-8} \text{ T}$ in the +z direction.

Magnetic Field on the Axis of a Circular Current Loop

Using the Biot-Savart law, we can calculate the magnetic field produced by a circular loop of wire carrying a steady current $I$. Let's find the field at a point P on the axis of the loop, at a distance $x$ from its center. The loop has a radius $R$.

By integrating the contributions from all the infinitesimal elements $dl$ around the loop, we find that the components of the magnetic field perpendicular to the axis cancel out due to symmetry. Only the components along the x-axis add up.

The magnetic field on the axis of a circular loop is: $\mathbf{B} = B_x \hat{\mathbf{i}} = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}} \hat{\mathbf{i}}$

A special, important case is the magnetic field right at the center of the loop, where $x=0$: $\mathbf{B}_0 = \frac{\mu_0 I}{2R} \hat{\mathbf{i}}$

The direction of the magnetic field can be found using another right-hand thumb rule:

• Curl the fingers of your right hand in the direction of the current flow in the loop.
• Your extended thumb points in the direction of the magnetic field at the center of the loop.

Given

• Number of turns, $N = 100$
• Radius, $R = 10 \text{ cm} = 0.1 \text{ m}$
• Current, $I = 1 \text{ A}$

To Find

The magnitude of the magnetic field, $B$, at the center.

Formula

For a single loop, $B = \frac{\mu_0 I}{2R}$. For N turns, the fields add up, so: $B = \frac{\mu_0 N I}{2R}$

Solution

Substitute the given values into the formula: $B = \frac{(4\pi \times 10^{-7} \text{ T m/A}) \times 100 \times 1 \text{ A}}{2 \times 0.1 \text{ m}}$ $B = \frac{4\pi \times 10^{-5}}{0.2} \text{ T} = 2\pi \times 10^{-4} \text{ T} \approx 6.28 \times 10^{-4} \text{ T}$

Final Answer The magnitude of the magnetic field at the center of the coil is $6.28 \times 10^{-4} \text{ T}$.

Ampere's Circuital Law

Ampere's Circuital Law provides an alternative way to calculate the magnetic field produced by a current. It is particularly useful in situations with high symmetry, much like Gauss's law is for electric fields.

The law relates the line integral of the magnetic field around a closed loop (called an Amperian loop) to the total current passing through the surface enclosed by that loop.

Ampere's Circuital Law is stated as: $\oint \mathbf{B} \cdot d\boldsymbol{l} = \mu_0 I_{\text{enclosed}}$

• The integral $\oint \mathbf{B} \cdot d\boldsymbol{l}$ is the line integral of the magnetic field around a closed path C.
• $I_{\text{enclosed}}$ is the total net current passing through the surface bounded by the path C.
• The right-hand rule determines the sign of the current: if you curl the fingers of your right hand in the direction of the integration path, your thumb points in the direction of positive current.

In simple, symmetric cases (like a long straight wire), the law can be simplified. If we choose an Amperian loop where the magnetic field B is tangential to the loop and has a constant magnitude $B$ along a length $L$, the law becomes: $BL = \mu_0 I_e$

Magnetic Field of a Long Straight Wire

Using Ampere's law, we can easily find the magnetic field at a distance $r$ from an infinitely long straight wire carrying current $I$. We choose a circular Amperian loop of radius $r$ centered on the wire.

• By symmetry, the magnetic field $B$ is constant in magnitude on this loop.
• The field B is tangential to the circle at every point, so $\mathbf{B}$ is parallel to $d\boldsymbol{l}$.
• The length of the loop is its circumference, $L = 2\pi r$.

Applying the simplified law: $B(2\pi r) = \mu_0 I$ $B = \frac{\mu_0 I}{2\pi r}$ This confirms Oersted's observation that the field gets weaker as you move away from the wire. The magnetic field lines are concentric circles around the wire.

The Solenoid

A solenoid is a coil consisting of a long wire wound in the form of a helix. When current flows through the wire, it creates a very useful magnetic field.

For a long solenoid (where the length is much greater than the radius), the magnetic field has the following properties:

• Inside the solenoid: The field is strong, uniform, and directed along the axis of the solenoid.
• Outside the solenoid: The field is very weak, approaching zero.

We can use Ampere's law to find the magnitude of the field inside. By choosing a rectangular Amperian loop, we can show that the magnetic field inside a long solenoid is: $B = \mu_0 n I$

• $\mu_0$ is the permeability of free space.
• $n$ is the number of turns per unit length ($n = N/L$).
• $I$ is the current in the wire.

Solenoids are widely used in science and technology to create controlled, uniform magnetic fields.

Given

• Length, $L = 0.5 \text{ m}$
• Number of turns, $N = 500$
• Current, $I = 5 \text{ A}$
• Radius, $r = 1 \text{ cm} = 0.01 \text{ m}$ (Note: The length $0.5$ m is much larger than the radius $0.01$ m, so we can use the long solenoid formula).

To Find

The magnitude of the magnetic field, $B$, inside the solenoid.

Formula

First, find the number of turns per unit length, $n$. $n = \frac{N}{L}$ Then, use the solenoid formula: $B = \mu_0 n I$

Solution

Calculate $n$: $n = \frac{500}{0.5 \text{ m}} = 1000 \text{ turns/m}$ Now calculate $B$: $B = (4\pi \times 10^{-7} \text{ T m/A}) \times (1000 \text{ m}^{-1}) \times (5 \text{ A})$ $B = 20\pi \times 10^{-4} \text{ T} \approx 62.8 \times 10^{-4} \text{ T} = 6.28 \times 10^{-3} \text{ T}$

Final Answer The magnitude of the magnetic field inside the solenoid is $6.28 \times 10^{-3} \text{ T}$.

Force between Two Parallel Currents, the Ampere

Since a current-carrying wire creates a magnetic field, and a wire in a magnetic field experiences a force, it follows that two current-carrying wires will exert forces on each other.

Consider two long, parallel wires 'a' and 'b', separated by a distance $d$, carrying currents $I_a$ and $I_b$.

1. Wire 'a' produces a magnetic field $B_a$ at the location of wire 'b'. The magnitude of this field is $B_a = \frac{\mu_0 I_a}{2\pi d}$.
2. Wire 'b' is now a current-carrying wire in the magnetic field of wire 'a'. It experiences a force. The force on a length $L$ of wire 'b' is $F_{ba} = I_b L B_a$.

Substituting the expression for $B_a$, the force on length $L$ of wire 'b' is: $F_{ba} = I_b L \left(\frac{\mu_0 I_a}{2\pi d}\right) = \frac{\mu_0 I_a I_b L}{2\pi d}$ The force per unit length, $f$, between the two wires is: $f = \frac{F_{ba}}{L} = \frac{\mu_0 I_a I_b}{2\pi d}$

By using the right-hand rules, we can determine the direction of this force. This leads to a simple rule:

• Parallel currents attract.
• Antiparallel (opposite direction) currents repel.

This is the opposite of electrostatics, where like charges repel.

Definition of the Ampere

This force between parallel wires is used to officially define the SI base unit of current, the ampere (A).

One ampere is the value of that steady current which, when maintained in each of two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to $2 \times 10^{-7}$ newtons per metre of length.

Torque on Current Loop, Magnetic Dipole

Torque on a Rectangular Current Loop

When a rectangular loop of wire carrying a current $I$ is placed in a uniform magnetic field B, it experiences a torque, but no net force.

The forces on the sides of the loop parallel to the field are zero or cancel out. The forces on the sides perpendicular to the field form a couple, which creates a torque that tends to rotate the loop.

The magnitude of this torque is: $\tau = IAB \sin\theta$

• $I$ is the current.
• $A$ is the area of the loop ($A=ab$).
• $B$ is the magnitude of the magnetic field.
• $\theta$ is the angle between the magnetic field B and the normal to the plane of the loop.

Magnetic Dipole Moment

We can define a vector quantity called the magnetic dipole moment, m, of the current loop: $\mathbf{m} = NI\mathbf{A}$

• $N$ is the number of turns in the coil.
• $I$ is the current.
• $\mathbf{A}$ is the area vector, whose magnitude is the area of the loop and whose direction is normal (perpendicular) to the plane of the loop, given by the right-hand rule.

Using the magnetic moment, the torque can be expressed elegantly as a vector product: $\boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}$ This is analogous to the torque on an electric dipole in an electric field: $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$.

The torque is zero when $\mathbf{m}$ is parallel or antiparallel to $\mathbf{B}$.

• Stable Equilibrium: When $\mathbf{m}$ is parallel to $\mathbf{B}$ ($\theta=0$). If displaced, the torque will restore it to this position.
• Unstable Equilibrium: When $\mathbf{m}$ is antiparallel to $\mathbf{B}$ ($\theta=180^\circ$). If slightly displaced, the torque will push it further away from this position.

Circular Current Loop as a Magnetic Dipole

At large distances, the magnetic field produced by a current loop is identical in form to the electric field produced by an electric dipole. This means a current loop acts as a magnetic dipole.

• On the axis of the loop (for distance $x \gg R$): $\mathbf{B} \approx \frac{\mu_0}{4\pi} \frac{2\mathbf{m}}{x^3}$
• This is analogous to the axial electric field of a dipole: $\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{2\mathbf{p}}{x^3}$.

A key difference is that an electric dipole is made of two separate electric charges (monopoles), but magnetic monopoles have never been observed. The most fundamental magnetic element known is the magnetic dipole, like a current loop.

Given

• Number of turns, $N = 100$
• Radius, $R = 10 \text{ cm} = 0.1 \text{ m}$
• Current, $I = 3.2 \text{ A}$
• Magnetic Field, $B = 2 \text{ T}$
• Moment of Inertia, $\mathscr{I} = 0.1 \text{ kg m}^2$

To Find

(a) Magnetic field at the center, $B_{\text{center}}$ (b) Magnetic moment, $m$ (c) Initial torque ($\tau_i$) and final torque ($\tau_f$) (d) Final angular speed, $\omega_f$

Formula

$B = \frac{\mu_0 N I}{2R}$ $m = NIA = NI(\pi R^2)$ $\tau = mB \sin\theta$ From rotational dynamics, work done by torque equals change in rotational kinetic energy: $\int \tau d\theta = \frac{1}{2}\mathscr{I}\omega_f^2$

Solution

(a) Calculate the magnetic field at the center Assuming $\pi \times 3.2 \approx 10$ as in the source for simplification. $B_{\text{center}} = \frac{(4\pi \times 10^{-7}) \times 100 \times 3.2}{2 \times 0.1} = \frac{4 \times 10^{-5} \times (\pi \times 3.2)}{0.2} \approx \frac{4 \times 10^{-5} \times 10}{0.2} = 2 \times 10^{-3} \text{ T}$ Answer for part (a) = $2 \times 10^{-3} \text{ T}$

(b) Calculate the magnetic moment $m = N I (\pi R^2) = 100 \times 3.2 \times \pi \times (0.1)^2 = 3.2\pi \approx 10 \text{ A m}^2$ Answer for part (b) = $10 \text{ A m}^2$

(c) Calculate the initial and final torques Initially, the axis of the coil (direction of $\mathbf{m}$) is in the direction of the field, so $\theta = 0^\circ$. $\tau_i = mB \sin(0^\circ) = 0$ Finally, the coil has rotated by $90^\circ$, so $\theta = 90^\circ$. $\tau_f = mB \sin(90^\circ) = (10 \text{ A m}^2)(2 \text{ T})(1) = 20 \text{ Nm}$ Answer for part (c) = Initial torque is $0$, final torque is $20 \text{ Nm}$.

(d) Calculate the final angular speed The work done by the torque as the coil rotates from $\theta=0$ to $\theta=\pi/2$ is converted into rotational kinetic energy. Work done $W = \int_{0}^{\pi/2} \tau d\theta = \int_{0}^{\pi/2} mB \sin\theta d\theta = mB [-\cos\theta]_{0}^{\pi/2} = mB(-\cos(90^\circ) - (-\cos(0^\circ))) = mB(0 - (-1)) = mB$. This work equals the final kinetic energy: $W = \frac{1}{2}\mathscr{I}\omega_f^2 \implies mB = \frac{1}{2}\mathscr{I}\omega_f^2$ $\omega_f = \sqrt{\frac{2mB}{\mathscr{I}}} = \sqrt{\frac{2 \times (10 \text{ A m}^2) \times (2 \text{ T})}{0.1 \text{ kg m}^2}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 \text{ s}^{-1}$ Answer for part (d) = $20 \text{ s}^{-1}$

The Moving Coil Galvanometer

The Moving Coil Galvanometer (MCG) is an instrument used to detect and measure small electric currents. Its operation is a direct application of the torque experienced by a current loop in a magnetic field.

Principle and Construction

• A coil with many turns is mounted on an axis so it can rotate freely.
• It is placed in a uniform radial magnetic field, usually created by concave pole pieces of a permanent magnet and a soft iron core placed inside the coil. A radial field ensures that the plane of the coil is always parallel to the magnetic field, meaning the torque is always maximum ($\sin\theta=1$).
• A spring is attached to the axis. When the coil rotates due to the magnetic torque, the spring provides a restoring counter-torque.

Working

1. When a current $I$ flows through the coil, it experiences a magnetic torque: $\tau = NIAB$.
2. This torque causes the coil to rotate.
3. As it rotates, the spring twists, creating a restoring torque proportional to the angle of deflection $\phi$: $\tau_{\text{restore}} = k\phi$, where $k$ is the torsional constant of the spring.
4. The coil stops rotating when the two torques are balanced: $NIAB = k\phi$
5. The deflection angle is therefore directly proportional to the current: $\phi = \left(\frac{NAB}{k}\right)I$ A pointer attached to the coil moves across a scale, allowing the current to be measured.

Galvanometer as an Ammeter

A galvanometer is too sensitive to measure large currents and has a significant resistance. To convert it into an ammeter (for measuring current in series), a very small resistance, called a shunt resistance ($r_s$), is connected in parallel with the galvanometer. Most of the current passes through the low-resistance shunt, protecting the galvanometer.

Galvanometer as a Voltmeter

To convert a galvanometer into a voltmeter (for measuring voltage in parallel), a very large resistance ($R$) is connected in series with it. This ensures that the voltmeter draws a negligible amount of current from the main circuit, so it doesn't significantly alter the voltage it is trying to measure.

Sensitivity of a Galvanometer

• Current Sensitivity: Deflection per unit current, $\frac{\phi}{I} = \frac{NAB}{k}$. It can be increased by increasing the number of turns, $N$.
• Voltage Sensitivity: Deflection per unit voltage, $\frac{\phi}{V} = \frac{NAB}{kR}$. Increasing the number of turns $N$ also increases the coil's resistance $R$, so doubling $N$ might double the current sensitivity but leave the voltage sensitivity unchanged.

Given

• Voltage source, $V = 3 \text{ V}$
• External resistance, $R = 3 \Omega$
• Galvanometer resistance, $R_G = 60.00 \Omega$
• Shunt resistance, $r_s = 0.02 \Omega$

To Find

The current $I$ in the circuit for three different measuring devices.

Formula

Ohm's Law: $I = \frac{V}{R_{\text{total}}}$ Resistance of ammeter (galvanometer + shunt in parallel): $R_{\text{ammeter}} = \frac{R_G r_s}{R_G + r_s}$

Solution

(a) Using the galvanometer directly The galvanometer is in series with the resistor. Total resistance $R_{\text{total}} = R_G + R = 60.00 \Omega + 3 \Omega = 63.00 \Omega$. $I = \frac{3 \text{ V}}{63.00 \Omega} \approx 0.048 \text{ A}$ Answer for part (a) = $0.048 \text{ A}$

(b) Using the galvanometer converted to an ammeter First, find the resistance of the ammeter (galvanometer with shunt). $R_{\text{ammeter}} = \frac{60.00 \Omega \times 0.02 \Omega}{60.00 \Omega + 0.02 \Omega} = \frac{1.2}{60.02} \Omega \approx 0.02 \Omega$ Now, find the total circuit resistance. $R_{\text{total}} = R_{\text{ammeter}} + R = 0.02 \Omega + 3 \Omega = 3.02 \Omega$. $I = \frac{3 \text{ V}}{3.02 \Omega} \approx 0.99 \text{ A}$ Answer for part (b) = $0.99 \text{ A}$

(c) Using an ideal ammeter An ideal ammeter has zero resistance. Total resistance $R_{\text{total}} = 0 + R = 3 \Omega$. $I = \frac{3 \text{ V}}{3 \Omega} = 1.00 \text{ A}$ Answer for part (c) = $1.00 \text{ A}$