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 $P_1$, $\vec{r} = j$

  $B = \frac{\mu_0}{4\pi} \frac{Idl \times r}{r^2}$

  $B = \frac{\mu<em>0}{4\pi} \frac{Idl (-i) \times j}{r^2} = \frac{\mu</em>0}{4\pi} \frac{Idl}{r^2} k$

  $B = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (125 A) (1 \times 10^{-2} m)}{4 \pi \times (1.2)^2}k$

  $B = -(8.7 \times 10^{-8} T)k$

  (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 $I<em>1$ and $I</em>2$ 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 $I<em>1$ becomes 10.0 A and $I</em>2$ 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:

  • $\mu_0 = 4 \pi \times 10^{-7} T \cdot m/A$
  • $I<em>1 = 5 A$, $I</em>2 = 2 A$
  • $L = 1.2 m$
  • $d = 0.4 m$

  $F = \frac{\mu<em>0 I</em>1 I_2 L}{2 \pi d}$

  $F = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (5 A) (2 A) (1.2 m)}{2 \pi \times 0.4 m} = 60 \times 10^{-7} N$

  (b) Given:

  • $I<em>1 = 10 A$, $I</em>2 = 4 A$

  $F = \frac{\mu<em>0 I</em>1 I_2 L}{2 \pi d}$

  $F = \frac{(4 \pi \times 10^{-7} T \cdot m/A) (10 A) (4 A) (1.2 m)}{2 \pi \times 0.4 m} = 2.4 \times 10^{-5} N$

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:

  $\oint \vec{B} \cdot d\vec{s} = \mu<em>0 I</em>{enc}$

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: