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In classical electromagnetism, Ampère's circuital law relates the circulation of a magnetic field around a closed loop to the electric current passing through the loop. This law is useful in finding the magnetic field due to currents under certain conditions of symmetry.

Ampere's Circuital Law

According to Ampere’s circuital equation, the line integral of a steady magnetic field across a closed loop is equal to μ0 times the total current (Ie) traveling through the surface bounded by the loop.

Ampere`s Law derivation from Biot- Savart`s Law Ampere's Law derivation from Biot- Savart's Law 

Applications of Ampere's Circuital Law

Magnetic Field Due to a Straight Infinite Current-Carrying Wire

  • Consider a current-carrying wire that is infinitely straight. We make an Amperian loop, which is a circle with a radius of r and a wire at its axis. As the magnetic field is tangential at all points on the loop, at all places, the magnetic field is of similar magnitude.
  • Suppose the magnetic induction at point P, distance R from the wire is required.

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

  • Draw the circle through P with center O and radius R as shown in the figure.
  • The magnetic induction Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced at all points along this circle will be the same and will be tangential to the circle, which is also the direction of the length element Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced
  • Thus, Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & AdvancedAmpere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced B × 2 πR. The current crossing of the circular area is i.
  • Thus, by Ampere's law, B × 2πR = μ0i

            ⇒ B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Note:

Line integral is independent of the shape of the path and the position of the wire within it.

The statement Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced does not necessarily mean that Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = 0 everywhere along the path but only that no net current is passing through the path.

Sign of current: The current due to which is produced in the same sense as (i.e. ) positive will be taken positive and the current which produces Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced in the sense opposite to Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced will be negative.

Example 1: Find the value of the loops L1, L2, and L3 in the figure shown. The sense of is mentioned in the figure.

Solution: for L1 Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

here I1 is taken positive because magnetic lines of force produced by I1 are clockwise as seen from the top. I2 produces lines of Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced in a clockwise sense as seen from the top. The sense of is anticlockwise as seen from the top.

for L2 : Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

for L3 : Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Special Case: To find out the magnetic field due to infinite current carrying wire

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

By B.S.L. Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advancedwill have circular lines.Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advancedis also taken tangent to the circle

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced Therefore, θ = 0° so Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = B 2πR (Therefore, B = const.)

Now by Ampere's law :

B 2πR = μ0I

Therefore, B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Hollow current carrying infinitely long cylinder : (I is uniformly distributed on the) whole circumference

(i) Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

By symmetry, the amperian loop is a circle.   

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced θ = 0

= Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

 Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced B = const.

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

(ii) r < R

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = μ0 (0) 

= B(2πr) = 0

Bin = 0 


Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Solid infinite current-carrying cylinder 

Assume current is uniformly distributed on the whole cross-section area

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Take an American loop inside the cylinder. By symmetry it should be a circle whose center is on the axis of the cylinder and its axis also coincides with the cylinder axis on the loop.

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = B . 2πr = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced ⇒ Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Case (II) :

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & AdvancedAmpere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = B.(2pr) = m0. I

⇒ B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced also Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Example 2: Consider a coaxial cable that consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c respectively. The inner wire carries an electric current i0 and the outer shell carries an equal current in the opposite direction. Find the magnetic field at a distance x from the axis where (a) x < a, (b) a < x < b (c) b < x < c and (d) x > c. Assume that the current density is uniform in the inner wire and also uniform in the outer shell.

Sol. Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

A cross-section of the cable is shown in the figure. Draw a circle of radius x with the center at the axis of the cable. The parts a, b, c, and d of the figure correspond to the four parts of the problem. By symmetry, the magnetic field at each point of a circle will have the same magnitude and will be tangential to it. The circulation of B along this circle is, therefore,

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

in each of the four parts of the figure.

(a) The current enclosed within the circle in part b is i0 so that

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Ampere's law

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced gives

B. 2πx = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced or, B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

The direction will be along the tangent to the circle.

(b) The current enclosed within the circle in part b is i0 so that

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

(c) The area of the cross-section of the outer shell is πc2 - πb2. The area of the cross-section of the outer shell within the circle in part c of the figure is πx2 - πb2.

Thus, the current through this part is Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced. This is in the opposite direction to the current i0 in the inner wire. Thus, the net current enclosed by the circle is

inet = i0 - Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced= Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

From Ampere's law,

B 2px = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced or, B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

(d) The net current enclosed by the circle in part d of the figure is zero and hence

B 2px = 0 or, B = 0.

Example 3(a): The figure shows a cross-section of a large metal sheet carrying an electric current along its surface. The current in a strip of width dl is Kdl where K is a constant. Find the magnetic field at a point P at a distance x from the metal sheet.

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Sol. Consider two strips A and C of the sheet situated symmetrically on the two sides of P (figure). The magnetic field at P due to strip A is B0 perpendicular to AP and that due to the strip C is Bc perpendicular to CP. The resultant of these two is parallel to the width AC of the sheet. The field due to the whole sheet will also be in this direction. Suppose this field has magnitude B.

Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

The field on the opposite side of the sheet at the same distance will also be B but in the opposite direction. Applying Ampere's law to the rectangle shown in the figure.

2B Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced = μ0 KAmpere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced or, B = Ampere`s Circuital Law & Its Applications | Physics for JEE Main & Advanced

Note that it is independent of x.

                

The document Ampere's Circuital Law & Its Applications | Physics for JEE Main & Advanced is a part of the JEE Course Physics for JEE Main & Advanced.
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FAQs on Ampere's Circuital Law & Its Applications - Physics for JEE Main & Advanced

1. What is Ampere's Law and how does it relate to electromagnetism?
Ans. Ampere's Law is a fundamental principle in electromagnetism that relates the magnetic field around a closed loop to the electric current flowing through the loop. It states that the integral of the magnetic field around a closed loop is equal to the product of the electric current passing through the loop and a constant known as the permeability of free space.
2. Can Ampere's Law be used to calculate the magnetic field of any current-carrying wire?
Ans. Yes, Ampere's Law can be used to calculate the magnetic field produced by any current-carrying wire, as long as the wire is symmetric and the magnetic field is constant along the chosen loop. By choosing an appropriate closed loop, the integral of the magnetic field can be evaluated to determine its value.
3. What are some practical applications of Ampere's Law?
Ans. Ampere's Law has various practical applications in different fields. It is used to analyze and design electromagnets, transformers, and electric motors. It also helps in understanding the behavior of magnetic materials and their interactions with electric currents. Additionally, Ampere's Law is essential in the field of electrical engineering for calculating magnetic fields in power transmission lines and determining the inductance of circuits.
4. How does Ampere's Law differ from Faraday's Law of electromagnetic induction?
Ans. Ampere's Law and Faraday's Law are two different principles in electromagnetism. Ampere's Law relates the magnetic field to the electric current, while Faraday's Law describes the induction of an electromotive force (emf) in a closed loop due to a changing magnetic field. Ampere's Law deals with the cause of magnetic fields, while Faraday's Law deals with the effect of changing magnetic fields on electric circuits.
5. Can Ampere's Law be used to calculate the magnetic field inside a current-carrying wire?
Ans. No, Ampere's Law cannot be used to directly calculate the magnetic field inside a current-carrying wire. Ampere's Law is applicable for closed loops outside the wire, where the magnetic field is easier to evaluate. To determine the magnetic field inside a wire, other techniques such as the Biot-Savart Law or symmetry considerations are used.
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