Ampere's Circuital Law & Its Applications | Physics Class 12 - NEET PDF Download

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 (I_e) traveling through the surface bounded by the loop.

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.

• Draw the circle through P with center O and radius R as shown in the figure.
• The magnetic induction  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
• Thus,  =  B × 2 πR. The current crossing of the circular area is i.
• Thus, by Ampere's law, B × 2πR = μ₀i

            ⇒ B =

Note:

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

The statement  does not necessarily mean that  = 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  in the sense opposite to  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 

here I₁ is taken positive because magnetic lines of force produced by I₁ are clockwise as seen from the top. I₂ produces lines of  in a clockwise sense as seen from the top. The sense of is anticlockwise as seen from the top.

for L₂ :

for L₃ :

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

By B.S.L. will have circular lines.is also taken tangent to the circle

 Therefore, θ = 0° so  = B 2πR (Therefore, B = const.)

Now by Ampere's law :

B 2πR = μ₀I

Therefore, B = 

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

(i) 

By symmetry, the amperian loop is a circle.   

 θ = 0

=

  B = const.

⇒ B = 

(ii) r < R

 =  = μ₀ (0) 

= B(2πr) = 0

⇒ B_in = 0 

Solid infinite current-carrying cylinder 

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

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.

 =  =  = B . 2πr =

B =  =  ⇒  =

Case (II) :

 =  = B.(2pr) = m₀. I

⇒ B =  also  =

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.    

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,

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 law

 gives

B. 2πx =  or, B =

The direction will be along the tangent to the circle.

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

(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 . This is in the opposite direction to the current i₀ in the inner wire. Thus, the net current enclosed by the circle is

i_net = i₀ - =

From Ampere's law,

B 2px =  or, B =

(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.

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 B_c 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.

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  = μ₀ K or, B =

Note that it is independent of x.