krit.club logo

Moving Charges and Magnetism - Ampere’s Circuital Law

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Ampere’s Circuital Law (ACL) relates the integrated magnetic field around a closed loop (called an Amperian loop) to the electric current passing through the loop. It states that the line integral of magnetic field B⃗\vec{B} around any closed path is equal to μ0\mu_0 times the total current IencI_{enc} threading through the loop: ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}.

Diagram showing a vertical wire with current I and a circular Amperian loop around it.
•

The Amperian loop is a mathematical construct, similar to a Gaussian surface. For a long straight wire, the magnetic field lines form concentric circles. By choosing a circular Amperian loop of radius rr centered on the wire, the magnetic field BB is constant in magnitude and tangent to the loop at every point, simplifying the integral to B(2πr)B(2\pi r).

Cross-section of a wire with a circular Amperian loop of radius r and the B field vector.
•

In an ideal long solenoid, the magnetic field is uniform and parallel to the axis inside, while it is approximately zero outside. Applying ACL to a rectangular loop (with one side inside and one side outside) shows that B=μ0nIB = \mu_0 n I, where nn is the number of turns per unit length.

Cross section of a solenoid with a rectangular Amperian loop used to calculate the internal magnetic field.
•

A toroid is essentially a solenoid bent into a circle to form a closed ring. The magnetic field is confined within the core of the toroid. According to ACL, for a loop of radius rr inside the core, the enclosed current is NINI, resulting in a field B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r}.

Diagram of a toroid showing the inner and outer radii and the Amperian loop within the core.

📐Formulae

∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}

B=μ0I2πr (Magnetic field due to a long straight wire)B = \frac{\mu_0 I}{2\pi r} \text{ (Magnetic field due to a long straight wire)}

B=μ0nI (Magnetic field inside a long solenoid)B = \mu_0 n I \text{ (Magnetic field inside a long solenoid)}

n=NL (Number of turns per unit length)n = \frac{N}{L} \text{ (Number of turns per unit length)}

B=μ0NI2πr (Magnetic field inside a toroid)B = \frac{\mu_0 N I}{2\pi r} \text{ (Magnetic field inside a toroid)}

💡Examples

Problem 1:

A long solenoid has 500500 turns per meter and carries a current of 2.0 A2.0\ A. Calculate the magnetic field at the center of the solenoid. Take μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7}\ T \cdot m/A.

Solution:

Given n=500 m−1n = 500\ m^{-1} and I=2.0 AI = 2.0\ A. Using the formula for a solenoid: B=μ0nIB = \mu_0 n I B=(4π×10−7)×500×2.0B = (4\pi \times 10^{-7}) \times 500 \times 2.0 B=4π×10−4 TB = 4\pi \times 10^{-4}\ T B≈1.26×10−3 TB \approx 1.26 \times 10^{-3}\ T.

Explanation:

The magnetic field inside a solenoid is directly proportional to the number of turns per unit length (nn) and the current (II). Since the solenoid is 'long', we use the formula for an ideal solenoid.

Problem 2:

A toroid has a core of inner radius 25 cm25\ cm and outer radius 26 cm26\ cm around which 35003500 turns of a wire are wound. If the current in the wire is 11 A11\ A, what is the magnetic field inside the core of the toroid?

Solution:

Mean radius r=25+262=25.5 cm=0.255 mr = \frac{25 + 26}{2} = 25.5\ cm = 0.255\ m. Total turns N=3500N = 3500, Current I=11 AI = 11\ A. Formula for toroid: B=μ0NI2πrB = \frac{\mu_0 N I}{2\pi r} B=4π×10−7×3500×112π×0.255B = \frac{4\pi \times 10^{-7} \times 3500 \times 11}{2\pi \times 0.255} B=2×10−7×3500×110.255B = \frac{2 \times 10^{-7} \times 3500 \times 11}{0.255} B≈3.02×10−2 TB \approx 3.02 \times 10^{-2}\ T.

Explanation:

For a toroid, the magnetic field is calculated using the mean radius of the circular path. The field exists only within the cross-section of the toroid core.

Problem 3:

A long straight solid conductor of radius RR carries a steady current II. The current is uniformly distributed across its cross-section. Use Ampere's Circuital Law to find the magnetic field at a distance rr where r<Rr < R.

Cross section of a solid conductor of radius R with an internal Amperian loop of radius r.

Solution:

  1. For r<Rr < R, the current enclosed IencI_{enc} is a fraction of the total current II based on the area ratio: Ienc=I×πr2πR2=Ir2R2I_{enc} = I \times \frac{\pi r^2}{\pi R^2} = I \frac{r^2}{R^2}
  2. Applying Ampere's Law: ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}
  3. B(2πr)=μ0(Ir2R2)B(2\pi r) = \mu_0 \left( I \frac{r^2}{R^2} \right)
  4. Solving for BB: B=μ0Ir2πR2B = \frac{\mu_0 I r}{2\pi R^2} This shows the field increases linearly with rr inside the conductor.

Explanation:

Inside a conductor with uniform current density, only the current within the cylinder of radius rr contributes to the magnetic field at that radius.

Problem 4:

A cylindrical cable consists of a thin inner wire and a concentric outer thin cylindrical shell of radius aa. The wire carries current II in one direction and the shell carries the same current II in the opposite direction. Find the magnetic field at a point outside the cable (r>ar > a).

Coaxial cable cross-section showing inner wire and outer shell with an Amperian loop enclosing both.

Solution:

  1. Choose a circular Amperian loop of radius r>ar > a centered on the cable.
  2. The total current enclosed by this loop is the algebraic sum of the currents: Itotal=Iinner+IouterI_{total} = I_{inner} + I_{outer}
  3. Since the currents are equal and opposite: Itotal=I+(−I)=0I_{total} = I + (-I) = 0
  4. Applying Ampere's Law: ∮B⃗⋅dl⃗=μ0(0)\oint \vec{B} \cdot d\vec{l} = \mu_0 (0) B(2πr)=0B(2\pi r) = 0 B=0B = 0 Thus, the magnetic field outside a coaxial cable with equal and opposite currents is zero.

Explanation:

Because the net current enclosed by the Amperian loop outside the cable is zero, the magnetic field in that region is also zero.