# 27.1 Magnets and Magnetic Fields #

- Every magnet has two ends or faces called
**poles**which are where the magnetic field is strongest- If a magnet is suspended so it can move freely, one pole will point north
- Aptly, this side is called the north pole

- Magnetic poles aren’t like electric charge: Positive or negative charge can easily be isolated, but we can never isolate a magnetic pole
- That is, if you cut a magnet is half you don’t obtain isolated north and south poles.
- Rather, you end up with two new magnets each with north and south poles

**Ferromagnetic:**Materials with a strong magnetic effect i.e. iron, cobalt, nickel, gadolinium- Similar to how we picture electric fields around a charge, we can picture magnetic fields surround a magnet
- Field lines should be drawn so that (1) the direction of the magnetic field is always tangent to a field line everywhere and (2) the number of lines per unit area is proportional to the magnetic field strength

(a) Visualizing magnetic field lines around a bar magnet, using iron filings and compass needles. The red end of the bar magnet is its north pole. The N pole of a nearby compass needle points away from the north pole of the magnet. (b) Diagram of magnetic field lines for a bar magnet.

## Earth’s Magnetic Field #

- Earth’s magnetic poles are not exactly through the geographic pole (axis of rotation)
- The angular difference between the direction of the compass needle (which points along the magnetic field lines) at any location and true (geographical) north varies between .$0 - 20^\circ$ with location
- Earth’s magnetic field at most location is not tangent to earth’s surface
**Angle of Dip:**The angle that Earth’s magnetic field makes with the horizontal at any point

The Earth acts like a huge magnet. But its magnetic poles are not at the geographic poles (on the Earth’s rotation axis).

# 27.2 Electric Currents Produce Magnet Fields #

(a) Deflection of compass needles near a current-carrying wire, showing the presence and direction of the magnetic field. (b) Iron filings also align along the direction of the magnetic field lines near a straight current-carrying wire. (c) Diagram of the magnetic field lines around an electric current in a straight wire. (d) Right-hand-rule-1 for remembering the direction of the magnetic field: when the thumb points in the direction of the conventional current, the fingers wrapped around the wire point in the direction of the magnetic field. (.$\vec B$ is the symbol for magnetic field).

# 27.3 Force on an Electric Current in a Magnetic Field #

- By Newton’s third law, we can see that a magnet exerts a force on a current-carrying wire
- The direction of the force is always perpendicular to the direction of the current and also perpendicular to the direction of the magnetic field .$\vec B$
- Use right hand rule! $$dF_\vec{B} = dq (\vec v \times \vec B) = dq\bigg(\frac{d\vec l}{dt}\times \vec B\bigg) = I (d\vec l \times \vec B)$$ $$\dots\ \vec F = I (\vec l \times \vec B) = I l b \sin\theta$$
- .$\vec l$ is the vector whose magnitude is the length of the wire its direction is along the (straight) wire in the direction of the conventional (positive) current
- We use the last equation if .$\vec B$ isn’t uniform or if the wire doesn’t form angle .$\theta$ with .$\vec B$ everywhere

# 27.4 Force on an Electric Charge Moving in a Magnetic Field #

- Recall, .$N$ particles, each charge .$q$, pass by a given point in time .$t$, they constitute current .$I = N q/t$
- Lets say in .$t$ time, a particle charge .$q$ moves distance .$l$ in a magnetic field .$\vec B$
- We know from kinematics that .$\vec l = \vec v t$ where .$\vec v$ is the velocity of the particle
- Using the 27.3 equation, we can find the force on all of these .$N$ particles as $$\vec F = I\vec l \times \vec B = (Nq/t)(\vec v t) \times \vec B = Nq\vec v \times \vec B$$
- Thus, the force on just
**one**of the .$N$ particles is $$\vec F = q \vec v \times \vec B = qvB \sin\theta$$

- Realize that
**we can save a lot of pain**if we know that the**magnetic field is uniform**in which case .$\vec F_\vec{B} = 0$ because the forces on opposite segments (sides) cancel out

## Uniform Field Path #

Force exerted by a uniform magnetic field on a moving charged particle (in this case, an electron) produces a circular path.

- Notice the field goes into the paper, denoted with .$\times$
- Because force is always orthogonal to .$\vec v$, the magnitude of .$\vec v$
- The centripetal acceleration has magnitude .$a = v^2/r$
- Thus we can derive $$F = ma \Longrightarrow qvB = m \frac{v^2}{r}$$ $$\dots \Longrightarrow r = \frac{mv}{qB}$$

- The time .$T$ it takes a particle with charge .$q$ and speed .$v$ to make a revolution is $$T = \frac{2\pi\cdot r}{v} = \frac{2\pi m}{qB}$$ $$f = \frac{1}{T} = \frac{qB}{2\pi m}$$

## Problem Solving #

Magnetic fields are somewhat analogous to the electric fields, but there are several important differences to recall:

- The force experienced by a charged particle moving in a magnetic field is
**orthogonal**to the direction of the magnetic field (and to the direction of the velocity of the particle), whereas the force exerted by an electric field is**parallel**to the direction of the field (and independent of the velocity of the particle). - The right hand rule, in its different forms, is intended to help you determine the direction of magnetic field, and the force a field exerts, and/or the directions of electric current or charged particle velocity. The right-hand rules to the right are designed to deal with the “perpendicular” nature of these quantities.

## With Electric Field #

**Lorentz Equation:**A particle charge .$q$ moving with velocity .$\vec v$ in the presence of both a magnetic field .$\vec B$ and electric field .$\vec E$ experiences a force $$\vec F = q (\vec E + \vec v \times \vec B)$$- Realize that the magnetic field cannot alter speed (do work), it can only alter the direction!
- To change an objects speed, you must apply a force
*along*the objects direction of motion - The magnetic field exerts a force on particles moving
*orthogonal*to it - Therefore, no work can be done because the particle can only move orthogonal to the magnetic field
- That being said, realize that this field is responsible for the circular, constant speed, motion
- This is why earth’s magnetic field deflects, but doesn’t slow down, charged particles from outer space

- To change an objects speed, you must apply a force

# 27.5 Torque on a Current Loop; Magnetic Dipole Moment #

Calculating the torque on a current loop in a magnetic field.$\vec B$

(a) Loop face parallel to .$\vec B$ field lines; (b) top view; (c) loop makes an angle to .$\vec B$, reducing the torque since the lever arm is reduced. The vector .$\vec \mu$ is the “magnetic moment”.

- When an electric current flows in a closed wire loop that’s in an external magnetic field, the magnetic force on the current can produce a torque
$$\tau = I aB \frac{b}{2} + I aB \frac{b}{2}$$
$$\dots = IabB$$
- .$A = ab$ is the area of the loop
- .$B$ is scalar of the magnetic field
- Notice, the vertical (orthogonal) sections of wire experience no force from the magnetic field

- If we have a coil of .$N$ loops of wire, the current is then .$NI$ so torque becomes
$$\tau = NIAB$$
- We call .$\vec \mu = NI \vec A$ the
**magnetic dipole moment** - The direction of .$\vec A$ (and thus .$\vec \mu$) is defined as perpendicular to the plane of the coil

- We call .$\vec \mu = NI \vec A$ the
- We can then re-write our torque eq as $$\vec \tau = NI \vec A \times \vec B$$ $$\dots \vec \tau = \vec \mu \times \vec B$$
- Dipoles have some potential energy, found by $$U = \int \tau d\theta = \int NIAB\sin\theta d \theta$$ $$\dots = -\mu B \cos\theta = - \vec \mu \cdot \vec B$$

# 27.8 Hall Effect #

- When a current-carrying conductor is held fixed in a magnetic field, the field exerts a sideways force on the charge moving in the conductor
- E.x, if electrons move to the right in the rectangular conductor, the inward magnetic field will exert a downward force: (a)
- This force is .$F_B = -e\vec v_d \times \vec B$
- .$v_d$ is drift velocity

- Thus, electrons tend to move towards side .$D$
- This creates a potential difference (called the
**Hall emf**), creating field .$\vec E_H$ - This field exerts force .$e\vec E_H$ on the moving charge

- This creates a potential difference (called the
- These forces are equal, that is, $$e E_H = ev_d B$$ $$\therefore E_H = v_d B$$
- Hall emf is then, asm uniform .$E_H$,
$$\mathscr{E}_H = E_H d = v_d B d$$
- .$d$ is the width of the conductor

The Hall effect. (a) Negative charges moving to the right as the current. (b) Positive charges moving to the left as the current.

# 27.9 Mass Spectrometer #

Mass spectrometers are used to measure the masses of atoms

Steps:

- Ions are produced (by a current or heating) and they pass through slit .$S_1$
- Ions then pass through a region with perpendicular electric and magnetic fields.
- Here, .$F_E = qE$ is equal to .$F_B = qvB$
- Therefore, .$v = \frac{E}{B}$ for all ions that pass the slit into .$S_2$; the rest are deflected

- Entering .$S_2$, there is only magnetic field .$B’$ so the ios follow a circular path
- Newtons second gives us .$F = ma \Longrightarrow qvB’ = mv^2/r$
- Since we know all terms, we can solve for mass: .$m = \frac{qB’r}{v} = \frac{qBB’r}{E}$