Unit – 4

The steady Magnetic Field and Magnetic Forces

Q1) Find the incremental field strength at P2 due to the current element of 2πa2 Am at P1. The co-ordinates of P1 and P2 are (4,0,0) and (0,3,0) respectively.

A1)

The two points P1 and P2 along with the I1dL1 current element at I1 are shown in figure.

According to Bio-Savart law

d   = I1 d x R12  / 4 π R 2 12   

= (0-4)   + (3-0) ay + 0. Az

R12  =       12  /|    12  |

= -4 x + 3   y   / 16 + 9

= -4 x + 3   y   /5

While I1 dL1 = 2   A m.

Q2) The following four vector fields are given in Cartesian co-ordinate system. The vector field, which does not satisfy the property of magnetic flux density, is

(a)

(c)

A2)

As we know that divergence of magnetic field is zero, therefore B=0

For the above relation becomes

From the given options, option(c) does not satisfy the above relation.

Hence the correct option is (c)

Q3) The plane y=1 carries current density Find H at A(0,0,0) and B(1,5,-2) .

A3)

The sheet is located at y=1 on which K is in direction. The sheet is infinite and shown in figure.

The will be in x direction.

a)     Point A(0,0,0)

  normal to current sheet at point A.

Q4) In the region 0<r<0.5m in cylindrical co-ordinates the current density is and  elsewhere. Use Amperes circuital law to find

A4)

The current from current density is given by

Q5) Consider a closed path with r ≥ 0.5 such that the enclosed current I is 1.876 A.

A5)

According to Ampere’ circuital law,

Q6) . Find at any point using Amper’s circuital law.

A6)

As current density is given by

Q7) A radial field   = 2.39 x 10 6 / r cos A/m exists in free space. Find the magnetic flux crossing the surface defined by 0 ≤≤ π/4 and 0 ≤ z ≤ 1m.

A7)

The portion of the cylinder is shown in figure. The flux crossing the given surface is given by

normal to direction is,

Q8) Find the flux passing the portion of the plane = π / 4 defined by 0.01 < r < 0.05m and 0<z<2 m. A current filament of 2.5 A is along z-axis in the direction in free surface.

A8)

Due to current carrying conductor in free space along z-axis

A/m

The flux crossing the surface is given by

normal to   direction

Q9) The plane of a circular coil is horizontal. It has 10 turns each of 8 cm radius. A current of 2 A flows through it, which appears clock wise from a point vertically above it. Find the magnitude and direction of the magnetic field at the centre of the coil due to current.

A9) Here, n=10, I=2A

r= 8cm = 0.08 m

Magnitude of magnetic field induction at the centre of the circular coil carrying current is

Q10) A long straight solid metal road of radius 4 cm carries a current 2A uniformly distributed over its circular cross section. Find the magnetic field induction of a distance 3 cm from the axis of wire.

A10)

The point P is lying inside the rod at perpendicular distance (r=3cm) from the axis of wire. Draw a circular closed path of radius r with centre on axis of rod such that the point P lies on this closed path.

Current enclosed by closed path

The magnetic field produced due to current flowing in the rod at every pointover the closed path is tangential to it and is equal in magnitude at all points of path.

Line integral of over closed path is

Or,

Q11) Determine the magnetic flux density B distance d meter from infinite straight wire carrying current I.

A11)

Let P be the point where B is to be evaluated consider an element it yr=dl of infinite wire shown in figure tattoo distance r from point P.

Figure: An infinite wire carrying a current I

Magnetic field B at P due to element dl from equation is given by

We see that

d

Therefore,

Substituting their values in above expression we have

Q12) Compute curl A if.

A12)

Given that

Partial derivatives will be

Putting these values of partial derivations in in equation

Q13) Verify that the vector fieldis both irrational (π has zero curl) and solenoid (π has zero divergence)

A13)

(i) given that

The partial derivatives will be

Putting these values of partial derivatives in curl equation

Hence the given field A is irrational since its curl zero

(ii) substituting the above calculated value of partial derivatives

Hence the given field is solenoid since divergence is zero.

Q14) Find the work done by the force vector .

Around the closed path as shown in figure.

A14)

The total work done by the force vector is

The integrals consists of 7 parts which can be calculated separately

This problem can be solved by Stokes theorem.

According two stokes theorem, the work done by force vector is

where S is surface bounded by the closed path and F is the given force vector for this force vector the curl is equal to

Choose normal vector

Q15) Find the curl of magnetic field and show that it is proportional to the current density.

Figure. Cross section of cylindrical conductor

A15)

For the cylindrical coordinates

But is entirely in the Q directions

The variation of is along r axis

Which is proportional to J.

Q16) Calculate the magnetic flux density due to coil 100 ampere turn and area of on the axis of coil at a distance 10 metre from the centre and at a point distance 10 metres in a direction at right angle to axis.

A16)

B on the axis of the coil

(ii) B ad perpendicular to axis

B at any point () is given by

M- Magnetic moment of a current loop and is equal to

When

Putting the given values we have

Q17) Circuit carrying a current of I ampere from a regular polygon of n side inscribed in circumscribing circle of radius R. Calculate the magnetic flux density at the centre of the polygon and show that for a circular loop if n tends to infinity.

A17)

Since the given regular polygon has n sides

P is perpendicular chopped from centre to the sides AB and OAB is an equilateral triangle.

B due to AB is given by

due to regular polygon of n sides will be

When n tends to

at the centre of the circular loop of radius R.

Figure. Regular polygon inscribed in a circle of radius R.

Q18) What is magnetic flux density?

A18) The magnetic flux density B is similar to the electric flux density D. As D = in free space, the magnetic flux density B is related to the magnetic field intensity H as:

B = ……..(1)

Where is a constant known as the permeability of free space. The constant is in henrys per meter (H/m) and has the value of

The magnetic Flux through a surface S is given by:

                                      …….(2)                  

Where the magnetic flux ' is in webers (Wb) and the magnetic flux density is in webers per square meter (Wb/m2) or Tesla’s (T).

Unlike electric flux lines, magnetic flux lines always close upon themselves. This is because it is not possible to have isolated magnetic poles (or magnetic charges).

                                  ……….(3)

This equation is referred to as the law of conservation of magnetic flux or Gauss's law for magnetostatic fields. Although the magnetostatic field is not conservative, magnetic flux is conserved.

By applying Divergence Theorem to equation 3, we obtain:

………..(4)

This equation is the fourth Maxwell's equation. It also shows that magnetostatic fields have no sources or sinks. It means that magnetic field lines are always continuous.

Q19) Explain amperes circuit law?

A19) Gauss’s Law is useful to obtain in case of complex problems. I magnetostatics complex problems can be solved using Ampere’s Law.

It states that

The line integral of magnetic field intensity around closed path is exactly equal to direct current enclosed by that path.

The mathematical representation of Ampere’s circuital law is

---------------------(1)

This law is useful to determine when current distribution is symmetrical.

Proof:

Consider a long straight conductor carrying direct current I placed along z axis as shown in figure. Consider a closed circular path of radius r which encloses the straight conductor carrying direct current I. The point P is at perpendicular distance r from the conductor.

Consider at point P which is direction direction tangential to circular path at point P.

Therefore,

While obtained at point P, from Biot-Savart law due to infinitely long conductor is,

= 1/ 2 π r . ɸ . r dɸ . ɸ

= 1/ 2 r . dɸ = 1/ 2 π . dɸ

Integrating . over the entire closed path

.   =   dɸ = 1/ 2 π [ɸ] 0 2π    = I. 2 π/ 2 π

This proves that the integral .   along closed path gives the direct current enclosed by that closed path.

      Steps to Apply Amper’s Circuital Law:

Step1: Consider a closed path preferably symmetrical such that it encloses the direct current I once. This is Amperian path. 

Step2: Consider differential length     depending upon the co-ordinate system used.

Step3: Identify the symmetry and find in which direction    exists according to co-ordinate system used.

Step 4: Find     .   . Make sure that   and      are in the same direction.

Step 5: Find the integral of .   around the closed path. And equate it to current I enclosed by the path.

To apply Ampere’s circuital law the following conditions must be satisfied,

1. The   ie either tangential or normal to the path, at each point of the closed path.
2. The magnitude of    must be same at all points of the path where   is tangential.

     due to straight conductors.

Consider an infinitely long straight conductor which is placed along z-axis carrying direct current I as shown in figure. Consider the Amperian closed path enclosing the conductor as shown in figure. Consider point P on the closed path at which is obtained. The radius of the path is r from the conductor. The magnitude of depends on r and the direction is always tangential to the closed path. So, has only component in direction say H .

Consider elementary length at point P in cylindrical co-ordinates it is r d in

   = Hɸ ɸ         and       = r dɸ . ɸ 

.   = Hɸ ɸ   .  r dɸ . ɸ   = Hɸ r dɸ

According to Ampere’s circuit law,

  = I

ɸ . r. dɸ = I

Hɸ . r  . dɸ = I

Hɸ . r  [2π] =I

Hɸ = 1/ 2π r

Hence   at point P is given by

= Hɸ. ɸ = 1/ 2 π r . ɸ

Q20) Find the H field within the slab when it is (a) permanently magnetized with magnetization Moi, (b) a linear permeable material with permeability A.

A20) For both cases, requires that the B field across the boundaries be continuous as it is normally incident. (a) For the permanently magnetized slab, this requires that

oH0 = o(H+M0)

H=H0-M0

Note that when there is no externally applied field (Ho = 0), the resulting field within the slab is oppositely directed to the magnetization so that B = 0.

(b) For a linear permeable medium requires

oH0 = o(H)

H= o/) H0

For > o the internal magnetic field is reduced. If H0 is set to zero, the magnetic field within the slab is also zero.