# Physics 12th Magnetic field due to current

- On 05/03/2013
- In PHYSICS

**Magnetic field due to current**

**(1) Introduction**

- In the previous chapter we have defined concept of magnetic field represented by vector
**B** - We defined magnetic field B in terms of force it exerts on moving charges and on current carrying conductors
- We also know that magnetic field is produced by the motion of the electric charges or electric current
- In this chapter we would study the magnetic field produced by the steady current
- we would study about how various factors affect the magnitude and direction of the magnetic field
- We will also learn to calculate the equation for magnetic field B if the current configuration is known using Biot-savart's law and ampere circuital law

**(2)Biot Savart Law**

- We know that electric current or moving charges are source of magnetic field
- A Small current carrying conductor of length
**dl**(length element ) carrying current I is a elementary source of magnetic field .The force on another similar conductor can be expressed conveniently in terms of magnetic field**dB**due to the first - The dependence of magnetic field
**dB**on current I ,on size and orientation of the length element**dl**and on distance**r**was first guessed by Biot and savart - The magnitude of the magnetic field
**dB**at a distance**r**from a current element**dl**carrying current I is found to be proportional to I ,to the length dl and inversely proportional to the square of the distance |**r**| - The direction of the magnetic Field is perpendicular to the line element
**dl**as well as radius**r** - Mathematically, Field
**dB**is written as

Here (μ_{0}/4π) is the proportionality constant such that

μ_{0}/4π=10^{-7}Tesla Meter/Ampere(Tm/A) - Figure below illustrates the relation between magnetic field and current element
- if in figure, Consider that line element
**dl**and radius vector**r**connecting line element mid point to the field point P at which field is to be found are in the plane of the paper - From equation (1) ,we expect magnetic field to be perpendicular to both
**dl**and**r**.Thus direction of**dB**is the direction of advance of right hand screw whose axis is perpendicular to the plane formed by**dl**and**r**and which is rotated from**dl**to**r**( right hand screw rule of vector product) - Thus in figure ,
**dB**at point P is perpendicular directed downwards represented by the symbol (x) and point Q field is directed in upward direction represented by the symbol (•) - The magnitude of magnetic field is

where θ is the angle between the line element dl and radius vector**r** - The resultant field at point P due to whole conductor can be found by integrating equation (1) over the length of the conductor i.e.
**B**=∫d**B****Relation between permeability (μ**_{0}and permittivity (ε_{0}) of the free space - We know that

μ_{0}/4π=10^{-7}N/A^{2}----(a)

and

1/4πε_{0}=9*10^{9}N-m^{2}/C^{2}----(b)

Dividing equation a by b we get

μ_{0}ε_{0}=1/(9*10^{16}) (C/Am)^{2}

we know that

1C=1A-s

So μ_{0}ε_{0}=1/(3*10^{8}m/s)^{2}

And 3*10^{8}m/s is the speed of the light in free space

So μ_{0}ε_{0}=1/c^{2}

or c=1/√(μ_{0}ε_{0})

**(3) Comparison between Coulomb’s laws and Biot Savart laws**

- Both the electric and magnetic field depends inversely on square of distance between the source and field point .Both of them are long range forces
- Charge element dq producing electric field is a scalar whereas the current element I
**dl**is a vector quantity having direction same as that of flow of current - According to coulomb’s law ,the magnitude of electric field at any point P depends only on the distance of the charge element from any point P .According to Biot savart law ,the direction of magnetic field is perpendicular to the current element as well as to the line joining the current element to the point P
- Both electric field and magnetic field are proportional to the source strength namely charge and current element respectively. This linearity makes it simple to find the field due to more complicated distribution of charge and current by superposing those due to elementary changes and current elements

**(4) Applications of Biot Savart law**

In this section we will now apply Biot-Savart law as studied in previous section to calculate field B in some important cases

**i) Magnetic Field due to steady current in an infinitely long straight wire**

- Consider a straight infinitely long wire carrying a steady current I
- We want to calculate magnetic field at a point P at a distance R from the wire as shown below in figure
- From Biot,-Savart law ,magnetic field d
**B**due to small current element of the wire at point O at a distance |**r**|=r from point P is

- since current element Id
**l**and vector**r**makes an angle θ with each other ,the magnitude of the product d**l**X**r**is dlrsinθ and is directed perpendicular to both dl and r vector as shown in the figure - Since from our choice of co-ordinate, we found out that field B lies along z-axis therefore we can write

where**k**is the unit vector along z-axis - we will now express sinθ and r in terms of R which is fixed distance for any point in space and l which describes the position of current element on the infinitely long wire .From figure 1 we have

and r=(R^{2}+l^{2})^{1/2}

Putting these values in the equation (5) we find - To find the field due to entire straight wire carrying wire ,we would have to integrate equation (6)
**B**=∫d**B** - To evaluate the integral on the RHS substitute

l=RtanΦ and dl=Rsec^{2}Φ dΦ

Therefore - From equation (7) ,we noticed that

i) Magnetic field is proportional to the current I

ii) It is inversely proportional to the distance R

iii)Magnetic field is in the direction perpendicular to the straight wire and vector**AP**=**R** - The magnetic line of force near a linear current carrying wire are concentric circles around the conductor in a plane perpendicular to the wire
- Hence the direction of field Bat point P at a distance R from wire, will be along the tangent drawn on a circle of radius R around the conductor as shown below in figure

**Direction of B**

- Direction of B can be found by right hand thumb rule i.e. grasp the wire with right hand ,the thumb pointing in direction of current ,the finger will curl around the wire in the direction of
**B** - The magnetic field lines are circular closed curve around the wire

**Application of Biot Savart law**

**ii) Force between two long and parallel current carrying conductor**

- It is experimentally established fact that two current carrying conductors attract each other when the current is in same direction and repel each other when the current are in opposite direction
- Figure below shows two long parallel wires separated by distance d and carrying currents I
_{1}and I_{2} - Consider fig 5(a) wire A will produce a field B
_{1}at all near by points .The magnitude of B_{1}due to current I_{1}at a distance d i.e. on wire b is

B_{1}=μ_{0}I_{1}/2πd ----(8) - According to the right hand rule the direction of
**B**is in downward as shown in figure (5a)_{1} - Consider length l of wire B and the force experienced by it will be (I
_{2}**l**X**B**) whose magnitude is

- Direction of F
_{2}can be determined using vector rule .F_{2}Lies in the plane of the wires and points to the left - From figure (5) we see that direction of force is towards A if I
_{2}is in same direction as I_{1}fig( 5a) and is away from A if I_{2}is flowing opposite to I_{1}(fig 5b) - Force per unit length of wire B is

- Similarly force per unit length of A due to current in B is

and is directed opposite to the force on B due to A. Thus the force on either conductor is proportional to the product of the current - We can now make a conclusion that the conductors attract each other if the currents are in the same direction and repel each other if currents are in opposite direction

**iii) Magnetic Field along axis of a circular current carrying coil**

- Let there be a circular coil of radius R and carrying current I. Let P be any point on the axis of a coil at a distance x from the center and which we have to find the field
- To calculate the field consider a current element Id
**l**at the top of the coil pointing perpendicular towards the reader - Current element Id
**l**and**r**is the vector joining current element and point P as shown below in the figure - From Biot Savart law, the magnitude of the magnetic field due to this current element at P is

where Φ is the angle between the length element dl and r - Since Id
**l**and**r**are perpendicular to each other so Φ=90.Therefore

- Resolving dB into two components we have dBsinθ along the axis of the loop and another one is dBcosθ at right angles to the x-axis
- Since coil is symmetrical about x-axis the contribution d
**B**due to the element on opposite side ( along -y axis ) will be equal in magnitude but opposite in direction and cancel out. Thus we only have dBsinθ component - The resultant B for the complete loop is given by,

B=∫dB

Now from figure 6

sinθ=R/r =R/√(R^{2}+ x^{2}) So

eq

- If the coil has N number of turns then

**Direction of B**

- Direction of magnetic field at a point on the axis of circular coil is along the axis and its orientation can be obtained by using right hand thumb rule .If the fingers are curled along the current, the stretched thumb will point towards the magnetic field
- Magnetic field will be out of the page for anti-clockwise current and into the page for clockwise direction

**Field at center of the coil**

- At the center of the coil x=0

so

**Field at point far away from the center x>>>R**

- In this case R in the denominator can be neglected hence

- For coil having N number of turns

- If the area of the coil is πR
^{2}then

- m=NIA represents the magnetic moment of the current coil. Thus from equation (17) we have

**iv) Magnetic Field at the center of a current carying arc**

- Consider an arc of radius R carrying current I as shown below in the figure
- According to the Biot Savart law the magnetic field at any point P is given by

Here dl=RdΦ

So

- If l is the length of the arc then

l=RΦ so that

- Equation 19 and 20 gives us magnetic field only at the center of curvature of a circular arc of current
- For semi circular loop put Φ=π in equation 19 and for full circle Φ=2π in equation 19 and calculate to find the result
- If the circular current loop lies on the plane of the paper then magnetic field will be out of the page for anticlockwise current and into the page for clockwise current as shown below in the figure

**(4) Ampere's circuital law**

- Ampere's circuital law in magnetism is analogous to gauss's law in electrostatics
- This law is also used to calculate the magnetic field due to any given current distribution
- This law states that

" The line integral of resultant magnetic field along a closed plane curve is equal to μ_{0}time the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant" Thus

where μ_{0}is the permeability of free space and I_{enc}is the net current enclosed by the loop as shown below in the figure - The circular sign in equation (21) means that scalar product
**B**.**dl**is to be integrated around the closed loop known as Amperian loop whose beginning and end point are same - Anticlockwise direction of integration as chosen in figure 9 is an arbitrary one we can also use clockwise direction of integration for our calculation depending on our convenience
- To apply the ampere's law we divide the loop into infinitesimal segments
**dl**and for each segment, we then calculate the scalar product of**B**and**dl** **B**in general varies from point to point so we must use**B**at each location of**dl**- Amperion Loop is usually an imaginary loop or curve ,which is constructed to permit the application of ampere's law to a specific situation

**Proof Of Ampere's Law**

- Consider a long straight conductor carrying current I perpendicular to the page in upward direction as shown below in the figure
- From Biot Savart law, the magnetic field at any point P which is at a distance R from the conductor is given by

- Direction of magnetic Field at point P is along the tangent to the circle of radius R withTh conductor at the center of the circle
- For every point on the circle magnetic field has same magnitude as given by

And field is tangent to the circle at each point - The line integral of B around the circle is

since ∫dl=2πR ie, circumference of the circle so,

This is the same result as stated by Ampere law - This ampere's law is true for any assembly of currents and for any closed curve though we have proved the result using a circular Amperean loop
- If the wire lies outside the amperion loop, the line integral of the field of that wire will be zero

but does not necessarily mean that**B**=0 everywhere along the path ,but only that no current is linked by the path - while choosing the path for integration ,we must keep in mind that point at which field is to be determined must lie on the path and the path must have enough symmetry so that the integral can be evaluated

**(5) Magnetic field of a solenoid**

- A solenoid is a long wire wound in a close-packed helix carrying a current I and the length of the solenoid is much greater then its diameter
- Figure below shows a section of a stretched out solenoid in xy and yz plane
- The solenoid magnetic field is the vector sum of the field produced by the individual turns that make up the solenoid
- Magnetic field B is nearly uniform and parallel to the axis of the solenoid at interior points near its center and external field near the center is very small
- Consider a dashed closed path abcd as shown in figure .Let l be the length of side ab of the loop which is parallel to the is of the solenoid
- Let us also consider that sides bc and da of the loop are very-very long so that side cd is very much far away from the solenoid and magnetic field at this side is negligibly small and for simplicity we consider its equal to 0
- At side a magnetic field
**B**is approximately parallel and constant. So for this side

∫B.dl=Bl - Magnetic field B is perpendicular to sides bc and da ,hence these portions of the loop does not make any contributions to the line integral as
**B**.**dl**=0 for the side bc and da - Side cd lies at external points solenoid where
**B**.**dl**=0 as B=0 or negligibly small outside the solenoid - Hence sum around the entire closed path reduces to Bl
- If N are number of turns per unit length in a solenoid then number of turns in length l is nl.The total current through the rectangle abcd is NIl and from ampere 's law

Bl=μ_{0}NlI

or B=μ_{0}NI (22) - we have obtained this relation for infinitely long solenoids considering the field at external points of the solenoid equal to zero.
- However for real solenoids external field is relatively weak rather then equal to zero
- Thus for actual solenoids relation 22 holds for internal points near the center of the solenoid
- Field at internal points of the solenoid does not depend on length and diameter of the solenoid and is uniform over the cross-section of a solenoid

**(6) Magnetic Field of a toriod**

- We will now apply Ampere circuital law to calculate magnetic field of a toriod
- A toriodal solenoid is a hollow circular ring with a large number of turns of a wire carrying current wound around the ring
- Suppose we have to find the magnetic field B at a point P inside the toriod as shown below in figure
- In this case amperion loop would be a circle through point P and concentric inside the toriod
- By symmetry field will have equal magnitude at all points of this circle and this field is tangential to every point in the circle

Thus

- If there are total N number of turns ,net current crossing the area bounded by the circle is NI where I is the current in the toriod
- using Ampere law

Thus we see that field B varies with r i.e. field B is not uniform over the cross-section of the core because the path l=2πr is longer at the outer side of the section then at the inner side - Imagine a concentric circle through point P
^{'}outside the toriod - The net current passing through this circular disc is zero ,since the current NI passes in and same current passes out. Thus using Ampere's circuital law, the field B=0 outside the torriod