Sources of Magnetic Fields

Biot-Savart law and Ampere's law for calculating magnetic fields

Sources of Magnetic Fields

Biot-Savart Law

Magnetic field from current element:

dB⃗=μ04πI dl⃗×r^r2d\vec{B} = \frac{\mu_0}{4\pi}\frac{I \, d\vec{l} \times \hat{r}}{r^2}

where μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} T·m/A is permeability of free space.

Total field: B⃗=μ0I4π∫dl⃗×r^r2\vec{B} = \frac{\mu_0 I}{4\pi}\int \frac{d\vec{l} \times \hat{r}}{r^2}

Infinite Straight Wire

Current II in straight wire:

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

Field circles wire by right-hand rule.

Circular Loop

On axis at distance xx from center, loop radius RR:

Bx=μ0IR22(x2+R2)3/2B_x = \frac{\mu_0 IR^2}{2(x^2 + R^2)^{3/2}}

At center (x=0x = 0): B=μ0I2RB = \frac{\mu_0 I}{2R}

Far from loop (x≫Rx \gg R): B≈μ0IπR22πx3=μ0μ2πx3B \approx \frac{\mu_0 I\pi R^2}{2\pi x^3} = \frac{\mu_0 \mu}{2\pi x^3}

where μ=IπR2\mu = I\pi R^2 is magnetic dipole moment.

Ampere's Law

∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

Line integral around closed path equals μ0\mu_0 times enclosed current.

Works best for:

  • Infinite straight wire
  • Solenoid
  • Toroid
  • Cylindrical symmetry

Long Solenoid

nn turns per unit length, current II:

Inside: B=μ0nIB = \mu_0 nI (uniform, parallel to axis)

Outside: B≈0B \approx 0

Toroid

NN total turns, inner radius aa, outer radius bb:

Inside toroid: B=μ0NI2πrB = \frac{\mu_0 NI}{2\pi r}

(varies with rr)

Outside: B=0B = 0

Magnetic Field of Moving Charge

Point charge qq moving with velocity v⃗\vec{v}:

B⃗=μ04πqv⃗×r^r2\vec{B} = \frac{\mu_0}{4\pi}\frac{q\vec{v} \times \hat{r}}{r^2}

Two Parallel Wires

Wires separated by distance dd, currents I1I_1 and I2I_2:

Force per unit length: FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1I_2}{2\pi d}

  • Same direction: attractive
  • Opposite direction: repulsive

This defines the ampere!

Magnetic Field Inside Conductor

For long straight conductor of radius RR:

Outside (r>Rr > R): B=μ0I/(2πr)B = \mu_0 I/(2\pi r)

Inside (r<Rr < R), uniform current density: B=μ0Ir2πR2B = \frac{\mu_0 Ir}{2\pi R^2}

(Linear in rr, like electric field in charged sphere)

Maxwell's Modification

With changing electric field:

∮B⃗⋅dl⃗=μ0Ienc+μ0ϵ0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{d\Phi_E}{dt}

The μ0ϵ0dΦE/dt\mu_0\epsilon_0 d\Phi_E/dt term is displacement current.

This completes Maxwell's equations!

📚 Practice Problems

1Problem 1medium

❓ Question:

A long straight wire carries current I = 15 A. Find the magnetic field at distances: (a) r = 2.0 cm from the wire, (b) r = 10 cm from the wire. (c) At what distance is the field B = 1.0 × 10⁻⁵ T? Use μ₀ = 4π × 10⁻⁷ T·m/A.

💡 Show Solution

Given:

  • I = 15 A
  • μ₀ = 4π × 10⁻⁷ T·m/A

(a) Field at r = 2.0 cm:

For long straight wire: B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

B=(4π×10−7)(15)2π(0.02)B = \frac{(4\pi \times 10^{-7})(15)}{2\pi(0.02)}

B=(2×10−7)(15)0.02=3.0×10−60.02B = \frac{(2 \times 10^{-7})(15)}{0.02} = \frac{3.0 \times 10^{-6}}{0.02}

B=1.5×10−4 T=0.15 mTB = \boxed{1.5 \times 10^{-4} \text{ T} = 0.15 \text{ mT}}

(b) Field at r = 10 cm:

B=(2×10−7)(15)0.10B = \frac{(2 \times 10^{-7})(15)}{0.10}

B=3.0×10−5 T=30 μTB = \boxed{3.0 \times 10^{-5} \text{ T} = 30 \text{ μT}}

(c) Distance for B = 1.0 × 10⁻⁵ T:

1.0×10−5=(2×10−7)(15)r1.0 \times 10^{-5} = \frac{(2 \times 10^{-7})(15)}{r}

r=(2×10−7)(15)1.0×10−5=3.0×10−61.0×10−5r = \frac{(2 \times 10^{-7})(15)}{1.0 \times 10^{-5}} = \frac{3.0 \times 10^{-6}}{1.0 \times 10^{-5}}

r=0.30 m=30 cmr = \boxed{0.30 \text{ m} = 30 \text{ cm}}

Note: B ∝ 1/r for long straight wire

2Problem 2hard

❓ Question:

A solenoid has N = 400 turns, length L = 0.25 m, and radius R = 0.02 m, carrying current I = 3.0 A. Find: (a) the magnetic field inside the solenoid (far from ends), (b) the magnetic field at the center of a circular coil with the same specifications, and (c) the self-inductance of the solenoid.

💡 Show Solution

Given:

  • N = 400 turns
  • L = 0.25 m
  • R = 0.02 m
  • I = 3.0 A
  • μ₀ = 4π × 10⁻⁷ T·m/A

(a) Field inside solenoid:

Turn density: n=N/L=400/0.25=1600n = N/L = 400/0.25 = 1600 turns/m

B=μ0nI=(4π×10−7)(1600)(3.0)B = \mu_0 n I = (4\pi \times 10^{-7})(1600)(3.0)

B=(4π×10−7)(4800)B = (4\pi \times 10^{-7})(4800)

B=6.03×10−3 T=6.03 mTB = \boxed{6.03 \times 10^{-3} \text{ T} = 6.03 \text{ mT}}

(b) Field at center of circular coil:

Using Biot-Savart law for N loops: Bcoil=μ0NI2RB_{coil} = \frac{\mu_0 N I}{2R}

Bcoil=(4π×10−7)(400)(3.0)2(0.02)B_{coil} = \frac{(4\pi \times 10^{-7})(400)(3.0)}{2(0.02)}

Bcoil=(4π×10−7)(1200)0.04B_{coil} = \frac{(4\pi \times 10^{-7})(1200)}{0.04}

Bcoil=3.77×10−2 T=37.7 mTB_{coil} = \boxed{3.77 \times 10^{-2} \text{ T} = 37.7 \text{ mT}}

Note: Coil field is stronger than solenoid field!

(c) Self-inductance:

L=μ0N2AlL = \frac{\mu_0 N^2 A}{l}

where A=πR2=π(0.02)2=1.26×10−3A = \pi R^2 = \pi(0.02)^2 = 1.26 \times 10^{-3} m²

L=(4π×10−7)(400)2(1.26×10−3)0.25L = \frac{(4\pi \times 10^{-7})(400)^2(1.26 \times 10^{-3})}{0.25}

L=(4π×10−7)(160000)(1.26×10−3)0.25L = \frac{(4\pi \times 10^{-7})(160000)(1.26 \times 10^{-3})}{0.25}

L=1.01×10−3 H=1.01 mHL = \boxed{1.01 \times 10^{-3} \text{ H} = 1.01 \text{ mH}}

3Problem 3hard

❓ Question:

Use Ampère's law to find the magnetic field: (a) inside a long cylindrical wire of radius R = 3.0 mm carrying uniform current density J = 5.0 × 10⁵ A/m², at radius r = 2.0 mm, and (b) outside the wire at r = 5.0 mm. (c) At what radius is the field maximum?

💡 Show Solution

Given:

  • R = 3.0 mm = 3.0 × 10⁻³ m
  • J = 5.0 × 10⁵ A/m² (uniform)
  • r₁ = 2.0 mm = 2.0 × 10⁻³ m (inside)
  • r₂ = 5.0 mm = 5.0 × 10⁻³ m (outside)

Total current: Itotal=J⋅πR2=(5.0×105)(π)(3.0×10−3)2=14.1 AI_{total} = J \cdot \pi R^2 = (5.0 \times 10^5)(\pi)(3.0 \times 10^{-3})^2 = 14.1 \text{ A}

(a) Field inside (r < R):

Current enclosed by circle of radius r: Ienc=J⋅πr2I_{enc} = J \cdot \pi r^2

Ampère's law: ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

B(2πr)=μ0Jπr2B(2\pi r) = \mu_0 J \pi r^2

B=μ0Jr2B = \frac{\mu_0 J r}{2}

At r = 2.0 mm: B=(4π×10−7)(5.0×105)(2.0×10−3)2B = \frac{(4\pi \times 10^{-7})(5.0 \times 10^5)(2.0 \times 10^{-3})}{2}

B=6.28×10−4 T=0.628 mTB = \boxed{6.28 \times 10^{-4} \text{ T} = 0.628 \text{ mT}}

(b) Field outside (r > R):

B=μ0Itotal2πrB = \frac{\mu_0 I_{total}}{2\pi r}

At r = 5.0 mm: B=(4π×10−7)(14.1)2π(5.0×10−3)B = \frac{(4\pi \times 10^{-7})(14.1)}{2\pi(5.0 \times 10^{-3})}

B=(2×10−7)(14.1)5.0×10−3B = \frac{(2 \times 10^{-7})(14.1)}{5.0 \times 10^{-3}}

B=5.64×10−4 T=0.564 mTB = \boxed{5.64 \times 10^{-4} \text{ T} = 0.564 \text{ mT}}

(c) Maximum field:

Inside: B=μ0Jr2B = \frac{\mu_0 Jr}{2} increases with r

Outside: B=μ0Itotal2πrB = \frac{\mu_0 I_{total}}{2\pi r} decreases with r

Maximum occurs at the surface:

rmax=R=3.0 mmr_{max} = \boxed{R = 3.0 \text{ mm}}

Bmax=μ0JR2=9.42×10−4 TB_{max} = \frac{\mu_0 JR}{2} = 9.42 \times 10^{-4} \text{ T}

🎴

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