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Moving Charges and Magnetism - Ampere’s Circuital Law

Grade 12CBSEPhysics

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

🔑Concepts

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Ampere’s Circuital Law states that the line integral of the magnetic field B⃗\vec{B} around any closed loop (called an Amperian loop) is equal to μ0\mu_0 times the net current II passing through the surface enclosed by the loop: ∮B⃗⋅dl⃗=μ0I\oint \vec{B} \cdot d\vec{l} = \mu_0 I.

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An Amperian loop is a closed path chosen such that at every point on the loop, B⃗\vec{B} is either tangential to the path with a constant magnitude or perpendicular to the path (making the dot product zero).

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For an infinitely long straight wire carrying current II, the magnetic field BB at a distance rr is derived using a circular Amperian loop of radius rr.

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A solenoid consists of a long wire wound in the form of a helix. For an ideal solenoid, the magnetic field inside is uniform and directed along its axis, while the field outside is considered zero.

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A toroid is essentially a solenoid bent into a circular shape. The magnetic field is confined within the core of the toroid and is zero in the open space inside and outside the toroid.

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The right-hand rule is used to determine the sign of the current: if the fingers of the right hand curl in the direction of the Amperian loop, the current in the direction of the thumb is taken as positive.

📐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.

Ampere’s Circuital Law - Revision Notes & Key Formulas | CBSE Class 12 Physics