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# Applications of Ampere's Theorem - Magnetism, Electromagnetic Theory, CSIR-NET Physical Sciences Physics Notes | EduRev

## Physics for IIT JAM, UGC - NET, CSIR NET

Created by: Akhilesh Thakur

## Physics : Applications of Ampere's Theorem - Magnetism, Electromagnetic Theory, CSIR-NET Physical Sciences Physics Notes | EduRev

The document Applications of Ampere's Theorem - Magnetism, Electromagnetic Theory, CSIR-NET Physical Sciences Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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Example: Magnetic Field Inside A Long Cylindrical Conductor

A cylindrical conductor with radius R carries a current I. The current is uniformly distributed over the cross-sectional area of the conductor. Find the magnetic field as a function of the distance r from the conductor axis for points both inside (r<R) and outside (r>R) the conductor.

From Ampereâ€™s Law, we have:

We will take the ampere loop to be a circle. Hence, for points inside the conductor, the ampere loop will be a circle with radius r, where r<R. The current enclosed will be

For points outside the conductor, the ampere loop will be a circle of radius r, where r>R . The current enclosed will just be I.

Example: Magnetic Field Of A Solenoid

A solenoid consists of a helical winding of wire on a cylinder, usually circular in cross section. If the solenoid is long in comparison with its cross-sectional diameter and the coils are tightly wound, the internal field near the midpoint of the solenoidâ€™s length is very nearly uniform over the cross section and parallel to the axis, and the external field near the midpoint is very small. Use Ampereâ€™s law to find the field at or near the center of such a long solenoid. The solenoid has n turns of wire per unit length and carries a current I.

From Ampereâ€™s Law, we have:

Following the integration path, we have:

Example: Magnetic Field Of A Toroidal Solenoid

The figure above shows a doughnut-shaped toroidal solenoid, wound with N turns of wire carrying a current I. Find the magnetic field at all points.

From Ampereâ€™s Law, we have:

Letâ€™s consider path 1,

No current is enclosed by the path. Hence, the magnetic field along the path is 0.

Letâ€™s consider path 3,

The net current enclosed by the path is 0. Hence, the magnetic field along the path is 0.

Letâ€™s consider path 2,

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