Magnetic Fields

Magnets and Magnetic Fields

Magnetic Poles

Magnets always come in pairs of North (N) and South (S) poles.
Magnetic monopoles (isolated N or S poles) have not been observed.

Force on Moving Charges

A magnetic field exerts a force on moving charges, which constitute currents.
The magnetic force is perpendicular to both the velocity of the charge and the magnetic field direction. This requires the use of the cross product in calculations.
Due to the perpendicular nature of the force, a moving charged particle in a uniform magnetic field will move in a circle or a spiral.

Force on Current-Carrying Wires

Because a moving charge is equivalent to a current, the force can be expressed in terms of current. However, since current is not a vector, this representation can be complex.

Magnetic Fields Created by Currents

Magnetic fields are created by electric currents.
Winding wire into a coil and running a current through it creates a magnet.
Zooming in on a permanent magnet reveals that the atomic charges are in motion, creating numerous tiny currents.
The strength of the magnetic field created by a current is proportional to the current and inversely proportional to the square of the distance from the current (1/r^2).

Right-Hand Rule for Magnetic Field

The direction of the magnetic field due to a moving positive charge can be determined using the right-hand rule.
- Point the thumb of your right hand in the direction of the velocity of the charge.
- Your fingers will curl in the direction of the magnetic field lines.
- If the charge is negative, the direction of the field lines is opposite to the direction indicated by the right-hand rule.
For field points, the velocity vector \vec{v} and the position vector \vec{r} both lie in the same plane.
The magnetic field \vec{B} is perpendicular to this plane.
The magnitude of the magnetic field at a point P is given by:
B = \frac{\mu_0 qv \sin{\theta}}{4 \pi r^2}
Where:
- \mu_0 is the permeability of free space.
- q is the charge of the particle.
- v is the velocity of the particle.
- \theta is the angle between the velocity vector and the position vector.
- r is the distance from the charge to the point P.

Magnetic Field of a Current Element

If there are n moving charged particles per unit volume, each with charge q, the total moving charge dQ in the segment is considered.

Biot-Savart Law

The magnetic field due to a current element is given by the Biot-Savart Law:
dB = \frac{\mu_0}{4 \pi} \frac{I dl \sin{\theta}}{r^2}
or in vector form:
d\vec{B} = \frac{\mu_0}{4 \pi} \frac{I d\vec{l} \times \hat{r}}{r^2}
Where:
- dB is the magnetic field due to the current element.
- \mu_0 is the permeability of free space.
- I is the current.
- dl is the length of the current element.
- r is the distance from the current element to the point where the field is being calculated.
- \hat{r} is the unit vector pointing from the current element to the point where the field is being calculated.
The magnetic field wraps in circles around a wire.
The direction of the magnetic field can be found using the right-hand rule: 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 B.
\mu_0 = permeability constant, approximately 4\pi \times 10^{-7} T \cdot m/A.

Magnetic Field due to a Long Straight Wire

The magnetic field due to a long straight wire can be calculated by summing up all the contributions ds to the current, keeping track of the distance r.
Given that r \sin(\theta) = R, the integral simplifies.
The magnetic field B due to the current in a long straight wire is:
B = \frac{\mu_0 I}{2 \pi R}
Where:
- R is the distance from the wire.

Magnetic Field at the Center of a Circular Arc of Wire

To calculate the magnetic field at the center of a circular arc of wire, sum up all the contributions ds to the current.
The distance r = R is constant.
For a complete loop, \phi = 2\pi:
B = \frac{\mu_0 I}{4 \pi R} \phi
For a full circle:
B = \frac{\mu_0 I}{2R}

Example Calculation: Magnetic Field from a Long Straight Conductor

A long, straight conductor carries a current of I = 1.0 A. Calculate the distance R from the axis of the conductor where the magnetic field has a magnitude of B = 0.5 \times 10^{-4} T.
Given:
- \mu_0 = 4 \pi \times 10^{-7} T \cdot m/A
- I = 1.0 A
- B = 0.5 \times 10^{-4} T
Using the formula:
B = \frac{\mu_0 I}{2 \pi R}
Solving for R:
R = \frac{\mu_0 I}{2 \pi B} = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (1.0 A)}{(2 \pi) (0.5 \times 10^{-4} T)} = 4 \times 10^{-3} m

Example: Magnetic Field of a Lightning Bolt

Lightning bolts can carry currents up to approximately 20 kA. Model such a current as the equivalent of a very long, straight wire.
- (a) Calculate the magnetic field at a distance of 5.0 m from the lightning bolt.
- (b) Compare this field to the field at 5.0 cm from a household current of 10 A.
(a) Given:
- \mu_0 = 4 \pi \times 10^{-7} T \cdot m/A
- I = 20 kA = 20 \times 10^3 A
- r = 5 m
B = \frac{\mu_0 I}{2 \pi r} = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (20 \times 10^3 A)}{2 \pi \times 5 m} = 8 \times 10^{-4} T
(b) Given:
- I = 10 A
- r = 5 cm = 5 \times 10^{-2} m
B = \frac{\mu_0 I}{2 \pi r} = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (10 A)}{2 \pi \times 5 \times 10^{-2} m} = 4 \times 10^{-5} T
Comparison: The magnetic field from the lightning bolt is 20 times larger than that from the household current.

Magnetic Field for Arcs

Example scenario on how to determine the magnetic field in the center of a loop of wire with lines and arcs.
If R = 10 cm and i = 2.43 A, and given arc angles of 95°, 90°, and 70°, calculate the magnetic field, considering that 95° = 1.658 radians, 90° = 1.571 radians, and 70° = 1.222 radians.

Example: Magnetic Field due to a Segment of Wire

A copper wire carries a steady 125-A current to an electroplating tank.
Find the magnetic field due to a 1.0-cm segment of this wire at a point 1.2 m away from it.
- (a) Point is straight out to the side of the segment.
- (b) Point is in the xy-plane and on a line at 30^\circ to the segment.
(a) At point P_1, \vec{r} = j
B = \frac{\mu_0}{4\pi} \frac{Idl \times r}{r^2}
B = \frac{\mu0}{4\pi} \frac{Idl (-i) \times j}{r^2} = \frac{\mu0}{4\pi} \frac{Idl}{r^2} k
B = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (125 A) (1 \times 10^{-2} m)}{4 \pi \times (1.2)^2}k
B = -(8.7 \times 10^{-8} T)k
(b)
B = \frac{\mu_0}{4\pi} \frac{Idl \times r}{r^2}
B = \frac{\mu_0}{4\pi} \frac{Idl (-i) \times (-i \cos(30^\circ) + j \sin(30^\circ))}{r^2}
Since i \times i = 0 and i \times j = k,
B = \frac{\mu_0}{4\pi} \frac{Idl \sin(30^\circ)}{r^2}k
B = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (125 A) (1.0 \times 10^{-2} m) \sin(30^\circ)}{4 \pi \times (1.2)^2}k
B = (-4.3 \times 10^{-8} T)k

Force Between Two Parallel Currents

A wire carrying a current in a magnetic field experiences a force.
When two parallel wires carry current, the magnetic field from one wire exerts a force on the other.
When the currents are parallel, the wires are pulled together (attractive force).
When the currents are anti-parallel, the wires are forced apart (repulsive force).
To calculate the force on wire b due to wire a:

Example: Force Between Superconducting Wires

Two straight, parallel, superconducting wires 4.5 mm apart carry equal currents of 15,000 A in opposite directions.
Calculate the force per unit length that each wire exerts on the other.
Given:
- \mu_0 = 4 \pi \times 10^{-7} T \cdot m/A
- I = 15000 A
- r = 4.5 \times 10^{-3} m
\frac{F}{L} = \frac{\mu_0 I^2}{2 \pi r} = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (15000 A)^2}{2 \pi (4.5 \times 10^{-3} m)} = 1.0 \times 10^4 Nm^{-1}

Example: Force Between Parallel Wires

Two long, parallel wires are separated by a distance of 0.400 m. The currents I1 and I2 have the directions as shown.
- (a) Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other. Is the force attractive or repulsive?
- (b) Each current is doubled, so that I1 becomes 10.0 A and I2 becomes 4.00 A. Now what is the magnitude of the force that each wire exerts on a 1.20-m length of the other?
(a) Given:
- \mu_0 = 4 \pi \times 10^{-7} T \cdot m/A
- I1 = 5 A, I2 = 2 A
- L = 1.2 m
- d = 0.4 m
F = \frac{\mu0 I1 I_2 L}{2 \pi d}
F = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (5 A) (2 A) (1.2 m)}{2 \pi \times 0.4 m} = 60 \times 10^{-7} N
(b) Given:
- I1 = 10 A, I2 = 4 A
F = \frac{\mu0 I1 I_2 L}{2 \pi d}
F = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (10 A) (4 A) (1.2 m)}{2 \pi \times 0.4 m} = 2.4 \times 10^{-5} N

Ampere’s Law

Ampere’s Law for magnetic fields is analogous to Gauss’ Law for electric fields.
Draw an “Amperian loop” around a system of currents.
The loop can be any shape, but it must be closed.
Sum the component of B along the loop, for each element of length ds around this closed loop.
The value of this integral is proportional to the current enclosed:
\oint \vec{B} \cdot d\vec{s} = \mu0 I{enc}

Magnetic Field Outside a Long Straight Wire with Current Using Ampere’s Law

For this case, B = \frac{\mu_0 I}{2 \pi r}.
Choose a circular Amperian loop (of radius r).
B and ds are parallel, and B is constant on the loop.

Magnetic Field Inside a Long Straight Wire with Current

Calculate B inside the wire.
The current is evenly distributed over the cross-section of the wire, it must be cylindrically symmetric.
Draw a circular Amperian loop around the axis, of radius r < R.
The enclosed current is less than the total current, because some is outside the Amperian loop. The amount enclosed is I_{enc} = I \frac{\pi r^2}{\pi R^2}.
So inside a straight wire:

Solenoids

A complete loop of wire has a magnetic field at its center.
Adding more loops makes the field stronger.
A many-turn coil of wire with current is called a solenoid.
Ampere’s Law can be used to calculate B inside the solenoid.
The field near the wires is circular, but farther away the fields blend into a nearly constant field down the axis.
Approximate that the field is constant inside and zero outside (just like capacitor).
Characterize the windings in terms of number of turns per unit length, n.
Each turn carries current i, so total current over length h is inh.

Toroids

The field of the solenoid sticks out both ends and spreads apart (weakens) at the ends.
Wrap the coil around like a doughnut, so that it has no ends. This is called a toroid.
The field has no ends, but wraps uniformly around in a circle.
Draw an Amperian loop parallel to the field, with radius r. If the coil has a total of N turns, then the Amperian loop encloses current Ni.
B inside:

Example: Solenoid Design

A solenoid is designed to produce a magnetic field of 0.0270 T at its center. It has radius 1.40 cm and length 40.0 cm, and the wire can carry a maximum current of 12.0 A.
- (a) What minimum number of turns per unit length must the solenoid have?
- (b) What total length of wire is required?
(a)
B = \mu_0 I n
n = \frac{B}{\mu_0 I} = \frac{0.027}{(4 \pi \times 10^{-7} T \cdot m/A) (12 A)} = 1790 turns/m
(b)
N = n \cdot l where l is the length.
First derive N from B = \frac{\mu_0 I N}{l}
From 2\pi r B= \mu IN where r is the radius:
(2π) (1.4 x 10-2) (2.7 x 10-2)
N = ----------------------------------= 158turns
(4л x 10-7)(12)
length of the wire = Nl = 157.5 x 0.4 = 63 m

Example: Magnetic Field at MIT

A magnetic field of 37.2 T has been achieved at the MIT Francis Bitter National Magnetic Laboratory.
Find the current needed to achieve such a field:
- (a) 2.00 cm from a long, straight wire
- (b) at the center of a circular coil of radius 42.0 cm that has 100 turns
- (c) near the center of a solenoid with radius 2.40 cm, length 32.0 cm, and 40,000 turns
(a)
B = \frac{\mu_0 I}{2 \pi R}
I = \frac{2 \pi R B}{\mu_0} = \frac{2 \pi (2 \times 10^{-2}) (37.2)}{4 \pi \times 10^{-7}} = 3.72 \times 10^7 A
(c) B = \mu_0 I n
n = \frac{N}{l} = \frac{40000}{0.32} = 125000 turns/m
I = \frac{B}{\mu_0 n} = \frac{37.2}{(4 \pi \times 10^{-7}) (1.25 \times 10^5)} = 237 A

Current-Carrying Coils

A current-carrying coil of wire acts like a small magnet.
The “dipole moment” (a vector) is defined as \vec{\mu} = NIA
The direction is given by the right-hand rule. Let your fingers curl around the loop in the direction of i, and your thumb points in the direction of \vec{B}.
The field lines of the loop look like they would if the loop were replaced by a magnet.
We can calculate the field in the center of such a loop but calculating other places is hard
But we can calculate along the z axis, where N is number of turns and A is area of loop.

B on Axis of Current-Carrying Coil

What is B at a point P on the z axis of the current loop?
Use the Biot-Savart Law to integrate around the current loop, noting that the field is perpendicular to r.
By symmetry, the perpendicular part of B is going to cancel around the loop, and only the parallel part will survive.

B Outside a Toroid

The magnetic field inside a Toroid is B = \frac{\mu_0 N i}{2 \pi r}.
What is the expression for the magnetic field outside?
Zero

Summary

Calculate the B field due to a current using Biot-Savart Law
Permeability constant: \mu_0
B due to long straight wire: or more accurately : the integral result B = \frac{\mu_0 I}{2 \pi R}.
circular arc: B = \frac{\mu_0 I}{4 \pi R} \phi
complete loop: B = \frac{\mu_0 I}{2R}
Force between two parallel currents
Another way to calculate B is using Ampere’s Law (integrate B around closed Amperian loops):
B inside a long straight wire:
a solenoid:
a torus:
B on axis of current-carrying coil: