Magnetic Field Due to A Current

Force per unit length: $\frac{dF}{dl} = \frac{\mu_0 i_1 i_2}{2\pi d}$

Derivation:

1. Wire 1 creates magnetic field $B_1 = \frac{\mu_0 i_1}{2\pi d}$ at the location of wire 2
2. Force on wire 2 is $F_2 = i_2 L \times B_1$
3. Force per unit length: $\frac{dF}{dl} = i_2 B_1 = i_2 \frac{\mu_0 i_1}{2\pi d} = \frac{\mu_0 i_1 i_2}{2\pi d}$

Direction:

• If currents are in same direction: wires ATTRACT each other
• If currents are in opposite directions: wires REPEL each other

Application: This principle defines the ampere. If two parallel wires 1m apart carry equal currents producing a force of $2 \times 10^{-7}$ N/m, the current is defined as 1 ampere.

$B = \frac{\mu_0 ia}{2\pi d\sqrt{a^2 + 4d^2}}$

Derivation:

1. Start with general expression from Biot-Savart Law for a finite wire segment: $B = \frac{\mu_0 i}{4\pi d}(\cos\theta_1 - \cos\theta_2)$

2. For a perpendicular bisector:

   • Wire segment length = $a$
   • Point is at distance $d$ from wire center
   • $\cos\theta_1 = \frac{a/2}{\sqrt{(a/2)^2 + d^2}} = \frac{a}{\sqrt{a^2 + 4d^2}}$
   • $\cos\theta_2 = -\frac{a}{\sqrt{a^2 + 4d^2}}$

3. Substituting: $B = \frac{\mu_0 i}{4\pi d}(2\frac{a}{\sqrt{a^2 + 4d^2}}) = \frac{\mu_0 ia}{2\pi d\sqrt{a^2 + 4d^2}}$

For a long wire ($a \gg d$), this approaches $B = \frac{\mu_0 i}{2\pi d}$ For a short wire ($a \ll d$), field decreases as $\frac{1}{d^2}$

The magnetic field at point P(0,d,0) is: $B = \frac{\mu_0 i}{4\pi d}\left(\sin\theta_2 - \sin\theta_1\right) = \frac{\mu_0 i}{4\pi d}\left(\frac{a}{\sqrt{a^2+4d^2}}\right)$

Analysis of field behavior:

1. When $a/d \ll 1$ (short wire segment):

   • $B \approx \frac{\mu_0 i}{4\pi}\frac{a}{d^2}$
   • Field falls off as $1/d^2$
   • Behaves like a magnetic dipole

2. When $a/d \approx 1$ (intermediate wire length):

   • Field transitions between dipole and long-wire behavior
   • Field strength is proportional to $\frac{\mu_0 i a}{d\sqrt{a^2+4d^2}}$

3. When $a/d \gg 1$ (long wire segment):

   • $B \approx \frac{\mu_0 i}{2\pi d}$
   • Field falls off as $1/d$
   • Approaches infinite wire case

Direction: The field at P is parallel to the z-axis (perpendicular to both the wire and the displacement vector).

Magnitude of magnetic field: $B = \frac{\mu_0}{4\pi} \frac{I \cdot dl \cdot \sin\theta}{r^2}$

$B = \frac{\mu_0}{4\pi} \frac{10 \text{ A} \cdot 10^{-2} \text{ m} \cdot \sin 45°}{(2 \text{ m})^2}$

$B = \frac{10^{-7} \text{ T·m/A} \cdot 10 \text{ A} \cdot 10^{-2} \text{ m} \cdot 0.707}{4 \text{ m}^2}$

$B \approx 1.8 \times 10^{-9} \text{ T}$

The approximation of treating the wire as a small element is valid because:

• The distance to the point (200 cm) is much larger than the length of the wire segment (1 cm)
• The ratio of distance to wire length is 200:1
• When this ratio exceeds about 10:1, the error from treating a finite wire as a point source becomes negligible
• This allows us to use the point-form of Biot-Savart Law rather than integrating over the length

The right-hand thumb rule states that:

• When you grip a straight current-carrying wire with your right hand, with your thumb pointing in the direction of conventional current
• Your curled fingers indicate the direction of the magnetic field lines around the wire

This creates circular magnetic field lines that encircle the wire. For example:

• If current flows upward through a wire, the magnetic field forms concentric circles around the wire in a counterclockwise direction when viewed from above
• At any point near the wire, the magnetic field direction is tangent to these circular field lines

Note: This rule is a direct consequence of the Biot-Savart law and illustrates the perpendicular relationship between current direction and magnetic field.

To determine the magnetic field at any point near a straight wire segment MN:

1. Identify geometric parameters:

   • Let d = perpendicular distance from point to wire
   • Determine angles $\theta_1$ and $\theta_2$ from the observation point to the wire ends
   • Calculate r (distance from current element to observation point)

2. Apply the general formula: $B = \frac{\mu_0 i}{4\pi d}(\cos\theta_1 - \cos\theta_2)$

3. For special cases:

   • For point on perpendicular bisector of wire with length a: $B = \frac{\mu_0 ia}{2\pi d\sqrt{a^2+4d^2}}$
   • For infinitely long wire: $B = \frac{\mu_0 i}{2\pi d}$

4. The field direction is always perpendicular to the plane containing the current element and the observation point, following the right-hand rule.

Example: For a 10 cm wire carrying 5A current, finding field at a point 2 cm away at an angle of 30° from one end requires calculating $\theta_1$ and $\theta_2$ before applying the formula.

When two parallel wires carry currents in the same direction:

1. They attract each other with a force per unit length given by: $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$ Where d is the separation distance between wires.

2. The physical explanation:

   • Each wire produces circular magnetic field lines
   • Wire 1 experiences a force due to being in Wire 2's magnetic field
   • Wire 2 experiences a force due to being in Wire 1's magnetic field
   • When currents flow in the same direction, each wire sits in a region where the other's field lines are oriented to produce an attractive force

3. This principle forms the basis for defining the ampere: two parallel wires 1m apart carrying 1A each experience a force of 2 × 10⁻⁷ N per meter.

If the currents flow in opposite directions, the wires would repel each other with the same magnitude of force.

The magnetic field at an axial point is:

$B = \frac{\mu_0 i a^2}{2(a^2 + d^2)^{3/2}}$

Where:

• $\mu_0$ is the permeability of vacuum ($4\pi \times 10^{-7}$ T·m/A)
• $i$ is the current in the loop
• $a$ is the radius of the loop
• $d$ is the distance from the center of the loop to the axial point

This formula shows that:

• The field is strongest at the center ($d = 0$)
• The field decreases with increasing distance $d$
• The direction of the field is along the axis of the loop

The magnetic field pattern evolves in the following way:

1. For separate circular loops: Each loop produces a dipole-like field that spreads outward in all directions
2. For adjacent coaxial loops: Fields between loops reinforce along the axis and partially cancel outside
3. For a loosely wound solenoid: Fields increasingly concentrate inside, with partial cancellation outside
4. For a tightly wound solenoid: Fields become nearly uniform inside and approach zero outside

This progression demonstrates the principle of field superposition, where the total magnetic field results from the vector addition of individual fields from each current loop.

To calculate required turns in a solenoid:

1. Start with the solenoid field equation from Ampere's Law: $B = \mu_0 ni$

2. Rearrange to solve for n (turns per unit length): $n = \frac{B}{\mu_0 i}$

3. For total number of turns N in a solenoid of length L: $N = n \times L = \frac{B \times L}{\mu_0 i}$

4. Calculation process:

   • Determine desired field strength B (in tesla)
   • Know the current i (in amperes)
   • Measure solenoid length L (in meters)
   • Use $\mu_0 = 4\pi \times 10^{-7}$ T·m/A

5. Example calculation: For B = 0.1 T, i = 2 A, L = 0.25 m: $N = \frac{0.1 \times 0.25}{4\pi \times 10^{-7} \times 2} = 9,947$ turns

The result gives the minimum turns needed, though practical factors like wire thickness and cooling may require additional considerations.