Ampere's Law and its Applications Class 12 Notes | EduRev

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Class 12 : Ampere's Law and its Applications Class 12 Notes | EduRev

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7. Ampere's law

This law is useful in finding the magnetic field due to currents under certain conditions of symmetry. Consider a closed plane curve enclosing some current-carrying conductors.

The line integral Ampere`s Law and its Applications Class 12 Notes | EduRev taken along this closed curve is equal to m0 times the total current crossing the area bounded by the curve. 

i.e.,    Ampere`s Law and its Applications Class 12 Notes | EduRev              ...(8)

where i = total current (algebraic sum) crossing the area.

As a simple application of this law, we can derive the magnetic induction due to a long straight wire carrying current i.

Suppose the magnetic induction at point P, distance R from the wire is required.

Ampere`s Law and its Applications Class 12 Notes | EduRev

Draw the circle through P with centre O and radius R as shown in figure.

The magnetic induction Ampere`s Law and its Applications Class 12 Notes | EduRev 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 Law and its Applications Class 12 Notes | EduRev

Thus, Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRevAmpere`s Law and its Applications Class 12 Notes | EduRev B × 2 πR

The current crossing the circular area is i.

Thus, by Ampere's law, B × 2πR = μ0i

⇒ B = Ampere`s Law and its Applications Class 12 Notes | EduRev

Note : -

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

The statement Ampere`s Law and its Applications Class 12 Notes | EduRev does not necessarily mean that Ampere`s Law and its Applications Class 12 Notes | EduRev = 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 Law and its Applications Class 12 Notes | EduRev in the sense opposite to Ampere`s Law and its Applications Class 12 Notes | EduRev will be negative.

Ex.11 Find the value of for the loops L1, L2, L3 in the figure shown. The sense of is mentioned in the figure.

Sol. for L1 Ampere`s Law and its Applications Class 12 Notes | EduRev

here I1 is taken positive because magnetic lines of force produced by I1 is anti clockwise as seen from top. I2 produces lines of Ampere`s Law and its Applications Class 12 Notes | EduRev in clockwise sense as seen from top. The sense of is anticlockwise as seen from top.

for L2Ampere`s Law and its Applications Class 12 Notes | EduRev

for L3Ampere`s Law and its Applications Class 12 Notes | EduRev

Ampere`s Law and its Applications Class 12 Notes | EduRev

7.1 Uses of Ampere's current law : To find out magnetic field due to infinite current carrying wire

Ampere`s Law and its Applications Class 12 Notes | EduRev Ampere`s Law and its Applications Class 12 Notes | EduRev

By B.S.L. Ampere`s Law and its Applications Class 12 Notes | EduRevwill have circular lines.Ampere`s Law and its Applications Class 12 Notes | EduRevis also taken tangent to the circle

Ampere`s Law and its Applications Class 12 Notes | EduRev Therefore, θ = 0° so Ampere`s Law and its Applications Class 12 Notes | EduRev = B 2πR (Therefore, B = const.)

Now by amperes law :

B 2πR = μ0I

Therefore, B = Ampere`s Law and its Applications Class 12 Notes | EduRev

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

(i) Ampere`s Law and its Applications Class 12 Notes | EduRev

By symmetry the amperian loop is a circle.   

Ampere`s Law and its Applications Class 12 Notes | EduRev

Ampere`s Law and its Applications Class 12 Notes | EduRev θ = 0

Ampere`s Law and its Applications Class 12 Notes | EduRev

 Ampere`s Law and its Applications Class 12 Notes | EduRev B = const.

Ampere`s Law and its Applications Class 12 Notes | EduRev

⇒ B = Ampere`s Law and its Applications Class 12 Notes | EduRev

(ii) r < R

Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev = μ0 (0) 

= B(2πr) = 0

⇒ Bin = 0 
Ampere`s Law and its Applications Class 12 Notes | EduRev

Graph

7.3 Solid infinite current carrying cylinder :

Assume current is uniformly distributed on the whole cross section area

Ampere`s Law and its Applications Class 12 Notes | EduRev

Ampere`s Law and its Applications Class 12 Notes | EduRev

taken an amperian loop inside the cylinder. By symmetry it should be a circle whose centre is on the axis of cylinder and its axis also coincides with the cylinder axis on the loop.

Ampere`s Law and its Applications Class 12 Notes | EduRev

Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev = B . 2πr = Ampere`s Law and its Applications Class 12 Notes | EduRev

B = Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev ⇒ Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev

Case (II) :

Ampere`s Law and its Applications Class 12 Notes | EduRev

Ampere`s Law and its Applications Class 12 Notes | EduRevAmpere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev = B.(2pr) = m0. I

⇒ B = Ampere`s Law and its Applications Class 12 Notes | EduRev also Ampere`s Law and its Applications Class 12 Notes | EduRev = Ampere`s Law and its Applications Class 12 Notes | EduRev

Ampere`s Law and its Applications Class 12 Notes | EduRev

Ex.12 Consider a coaxial cable which 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 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 Law and its Applications Class 12 Notes | EduRev Ampere`s Law and its Applications Class 12 Notes | EduRev Ampere`s Law and its Applications Class 12 Notes | EduRev Ampere`s Law and its Applications Class 12 Notes | EduRev

A cross-section of the cable is shown in figure. Draw a circle of radius x with the centre 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 Law and its Applications Class 12 Notes | EduRev

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 and its Applications Class 12 Notes | EduRev

Ampere's law

Ampere`s Law and its Applications Class 12 Notes | EduRev gives

B. 2πx = Ampere`s Law and its Applications Class 12 Notes | EduRev or, B = Ampere`s Law and its Applications Class 12 Notes | EduRev

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 Law and its Applications Class 12 Notes | EduRev

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

Thus, the current through this part is Ampere`s Law and its Applications Class 12 Notes | EduRev. This is in the opposite direction to the current i0 in the inner wire. Thus, the net current enclosed by the circle is

inet = i0Ampere`s Law and its Applications Class 12 Notes | EduRevAmpere`s Law and its Applications Class 12 Notes | EduRev

From Ampere's law,

B 2px = Ampere`s Law and its Applications Class 12 Notes | EduRev or, B = Ampere`s Law and its Applications Class 12 Notes | EduRev

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

B 2px = 0 or, B = 0.

Ex.13(a)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 Law and its Applications Class 12 Notes | EduRev

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 the 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 Law and its Applications Class 12 Notes | EduRev Ampere`s Law and its Applications Class 12 Notes | EduRev

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

2B Ampere`s Law and its Applications Class 12 Notes | EduRev = μ0 KAmpere`s Law and its Applications Class 12 Notes | EduRev or, B = Ampere`s Law and its Applications Class 12 Notes | EduRev

Note that it is independent of x.

Ex.13 (b) A cylinder of radius R1 have cylindrical cavity of radius R2 as shown in the figure and have current density J (down ward). Find Magnetic field when

(i) r > R1 (on x-axis towards right)

(ii) r < R1 (on x-axis towards left)
Ampere`s Law and its Applications Class 12 Notes | EduRev

(iii) when point is in cavity

Sol. (i) r > R1 (on x-axis towards right)

Ampere`s Law and its Applications Class 12 Notes | EduRev

Magnetic field due to big cylinder = B1

Magnetic field due to smaller cylinder = B2

Br = B1 + B2

Ampere`s Law and its Applications Class 12 Notes | EduRev

(ii) r < R1 (on x-axis) towards left of centre Magnetic field due to big cylinder

B1Ampere`s Law and its Applications Class 12 Notes | EduRev

Magnetic field due to smaller cylinder 

B2Ampere`s Law and its Applications Class 12 Notes | EduRev

So net Magnetic field                            

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