Magnetic Field

Magnetic Field

Magnetic Field

Orested Discovery

Orested took a magnetic compass near to an electric circuit and observed deflection on the magnetic compass when current flows through the circuit as shown in figure(b). The direction of deflection is reversed when the terminal of the circuit is reversed as shown in the figure (c). On the basis of this experiment Orested concluded that a magnetic field is induced in the vicinity of current carrying conductor. This discovery is called Orested discovery.

Direction of induced magnetic field

The direction of induced magnetic field is explained by following two rules.

  1. Right hand thumb rule

    According to this rule," If the stretched thumb of right hand represents the direction of electric current, the curved fingers of same hand represent the direction of induced magnetic field."

  2. Right hand fist/curl/grip rule

    According to this rule,"If the curved fingers of right hand represents the direction of electric current , the stretched thumb of same hand represents the direction of induced magnetic field."

Lorentz Force

The forced experienced by a charge particle moving in the magnetic field is called Lorentz Magnetic Force and the force experienced by a charged particles placed in electric field is called Lorentz Electric Force. The sum of Lorentz Magnetic Force and Lorentz Electric Force on a charge particle is called Lorentz Force after the name of Dutch Scientist Lorentz who formulated it.

Force Experienced by a Charged Particle Moving in Uniform Magnetic Field

Let us consider a charged particle with charge 'q' is moving in a uniform magnetic field 'B' with velocity of 'v' by making angle of /theta with the magnetic field. Then, the Lorentz Magnetic Force it experiences is,

  • Directly proportional to the strength of magnetic field
    F
    α B ............(i)
  • Directly proportional to the magnitude of charge
    F
    α q ............(ii)
  • Directly proportional to velocity of particle
    F
    α B ............(iii)
  • Directly proportional to the sine of angle between v and B
    F
    α Sin θ ............(iv)

From equation (i), (ii) , (iii) and (iv),
F
α B q v Sin θ
or, F = k B q v Sin θ
where k is a proportionality constant with value 1.
F = B q v Sin θ
In vector form,
\(\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})\)

Special Cases:

  • When θ = 0° or 180°
    F = B q v Sin 0°
    = 0
    F= B q v Sin 180
    °
    = 0
  • When θ = 90°
    F = B q v Sin 90°
    = B q v

Force experienced by a straight current carrying conductor placed in uniform magnetic field

Let us consider a straight conductor of length 'l' ,cross sectional area 'A' carrying current 'i' is placed in a uniform magnetic field of strength 'B'. Let θ be the angle made by conductor with magnetic field as shown in figure.Then,

Volume of conductor = A .l

If n be the number of free electrons(es) per unit volume of the conductor,

Total number of free electrons(es) in conductor = n A l

Force experienced by a electron moving in uniform magnetic field = B e v Sin θ

Force experienced by nAl electrons moving in uniform magnetic field = nAl Bev Sin θ

Since, total forced experienced by all free electrons is total force experienced by the conductor.
Hence, F = n A l B e v Sin θ ............(i)

But, i = v e n a ............(ii)

From equation (i) and (ii) we get,

F = B i l Sin θ

This is the required expression. In vector form,
\(\overrightarrow{F} = i \times (\overrightarrow{l} \times \overrightarrow{b})\)

The direction of Lorentz force is given by Flemings Left Hand Rule which states that, "The first three finger's in left hand held mutually perpendicular to each other such that the index finger shows the direction of magnetic field, the middle finger shows the direction of current and thumb shows the direction of Lorentz force.

Torque experienced by a rectangular current carrying coil when placed in uniform magnetic field

Let us consider a rectangular coil PQRS of length 'l' , bredth 'b' and area 'A' carrying current 'i' in anti clock wise direction is placed in uniform magnetic field of strength B as shown in figure.

The total force experienced by the coil is the resultant of Lorentz force experienced by all arms of the coil. Now,

Force experienced by PQ(FPQ) = B i l Sin θ
But, θ = 90
°
= B i l
According to Fleming's Left Hand Rule, FPQ acts normally outward the paper as shown in the figure.

Force experienced by QR(FQR) = B i b Sin θ
According to Fleming's Left Hand Rule, FPQ acts normally downward the paper as shown in the figure.

Force experienced by RS(FRS) = B i l
According to Fleming's Left Hand Rule, FRS acts normally inward the paper as shown in the figure.

Force experienced by PS(FPS) = B i b Sin θ
According to Fleming's Left Hand Rule, FPS acts normally upward the paper as shown in the figure.

Force FPS and FQR are equal and act in opposite direction in same line. Hence, they balance each other. However, FPQ and FRS generates torque on the coil.

Now,
Torque(
τ) = Force × perpendicular distance between PQ and RS
i.e
τ = B i l . PA ............(i)

In Δ PAS
\(\cos \theta = \frac{PA}{PS}\)
or, PA = b Cos θ  ............(ii)

From (i) and (ii)
τ = B i l . b cos θ
or,
τ = B i A cos θ

If coil consist of N no. of turns,
τ = B i N A Cos θ
which is the required expression.

Moving Coil Galvanometer(MCG)

A galvanometer is a device which is used to detect the current in an electric circuit.

Principle:
A moving coil galvanometer is based on the principle that when an electric current is passed through a rectangular coil placed in a uniform magnetic field , the coil experience a torque that produce deflection on the coil.

Construction
A moving coil galvanometer consist of a rectangular coil with soft magnetic material at its core placed in a uniform magnetic field between two concave poles of magnet. The coil is suspended by a phosphor bronze stripe to the torsion head. Lower part of the coil is connected with a hair spring. A concave mirror is placed on the phosphor bronze stripe and a lightening system is used to detect the deflection. Terminal T1 is connected to the phosphor bronze stripe and terminal T2 is connected to the hair spring through which electric current is passed to the coil as shown in the figure.

Theory
When current 'i' is passed to the coil's two terminals T1 and T2, the coil experiences a torque given by,
Torque due to deflection(
τd) = B i N A Cos θ ............(i)

Since we are using the concave poles of magnet, i.e the magnetic field is radial. Therefore, θ = 0°
Hence, τd = B i N A

The restoring torque(τr) = k α
Where, k = spring constant , α = angle of deflection

At equilibrium,
τd = τr
i.e B i N A = k α
or, \(i = \frac{k \alpha}{B N A}\) ............(iii)

From eqn (iii), k, B, N and A are constant. Hence, i is directly proportional to α i.e MCG has linear scale.

Current Sensitivity

The deflection per unit current on MCG is called current sensitivity.
Mathematically, \(Current\ sensitivity(S_{i}) = \frac{Deflection(\alpha)}{Current(i)}\)
or, \(S_{i} = \frac{\alpha}{i}\) ............(iv)

From equation (iii) and (iv),
\(S_{i} = \frac{\alpha}{(\frac{k \alpha}{B N A})}\)
or, \(S_{i} = \frac{\alpha}{k \alpha} B N A \)
or, \(S_{i} = \frac{B N A}{k}\) ............(v)

Voltage Sensitivity

The deflection per unit voltage on MCG is called voltage sensitivity.
Mathematically, \(Current\ sensitivity(S_{i} = \frac{Deflection(\alpha)}{Voltage(V)}\)
or, \(S_{i} = \frac{\alpha}{i R}\) ............(vi)

From equation (iii) and (vi),
\(S_{i} = \frac{\alpha}{(\frac{k \alpha}{B N A}) R}\)
or, \(S_{i} = \frac{\alpha}{k \alpha} \frac{B N A}{R} \)
or, \(S_{i} = \frac{B N A}{k R}\)

Ways to increase current sensitivity of a MCG

  1. By increasing number of turns.
  2. By increasing strength of magnetic field induced.
  3. By decreasing cross sectional area.

Biot Savart's Law

Biot Savart's Law gives the quantative measurement of magnetic field induced due to current flowing on the conductor. According to Biot Savart's Law,  The small magnetic field induced at a point at a distance 'r' from current element 'dl' carrying current 'i' is,
\(dB = \frac{\mu_{o}}{4 \pi} \frac{i dl \sin \theta}{r^{2}}\)
where
μo is the perneability of free space or vaccum.

Explanation

The small magnetic field(dB) induced at point P is,

  • Directly proportional to the current flowing in the conductor,
    i.e dB
    α i ............(i)
  • Directly proportional to the length of current element,
    i.e dB
    α dl ............(ii)
  • Directly proportional to the sine of angle between radius vector and current element,
    i.e dB
    α Sin θ ............(iii)
  • Inversely proportional to the square of radius vector,
    i.e dB
    α 1\r2 ............(iv)

Combining equation (i),(ii),(iii) and (iv),
\(dB \propto \frac{i dl \sin \theta}{r^{2}}\)
or, \(dB = \frac{\mu_{o}}{4 \pi} \frac{i dl \sin \theta}{r^{2}}\)

where, \(\frac{\mu_{o}}{4 \pi}\) is proportionality constant.

Application of Biot Savart's Law

  1. To calculate the magnetic field induced at a center of current carrying circular coil
    Let us consider a circular coil of radius 'r' carrying current 'i' in anti clockwise direction as shown in the figure. We are intrested to calculate the magnetic field induced at the center (P) as shown in the figure.
    Let us take a small current element 'dl' on the circumference of the circle due to which 'dB' magnetic field is induced at the center of the coil.
    Now According to Biot Savart's Law,
    \(dB = \frac{\mu_{o}}{4 \pi} \frac{i dl \sin \theta}{r^{2}}\)
    But θ = 90
    ° (Since, Angle between tangent and radius vector)
    or, \(dB = \frac{\mu_{o}}{4 \pi} \frac{i dl }{r^{2}}\)

    \(B = \int_{0}^{2 \pi r} dB\)
    \(= \int_{0}^{2 \pi r} \frac{\mu_{o}}{4 \pi} \frac{i dl }{r^{2}}\)
    \(= \frac{\mu_{o}i}{4 \pi r^{2}} \int_{0}^{2 \pi r} dl\)
    \(= \frac{\mu_{o}i}{4 \pi r^{2}} [l]^{2 \pi r}_{0}\)
    \(=\frac{\mu_{o}i}{4 \pi r^{2}} \times 2 \pi r\)
    \(=\frac{\mu_{o}i}{2r}\)
    If coil consist of "N" no. of turns,
    \(B = \frac{\mu_{o}Ni}{2r}\)

    Direction: The direction of induced magnetic field is normally outward the paper according to right hand curl rule.

  2. Magnetic field induced at a point on the axis of current carrying circular coil
    Let us consider a circular coil of radius 'a' carrying current 'i' in anti clockwise direction. We are intrested to calculate the magnetic field induced at a point 'P' on the axis of the coil.
    • dB cos φ along vertical
    • dB sin φ along horizontal , as shown in the figure.
    Let us take small current element 'dl' on the circumference of the coil due to which dB magnetic field is induced at point P that lies 'x' distance away from the center of the coil.
    According to Biot Savart's Law,
    \(dB = \frac{\mu_{o}}{4 \pi} \frac{i dl \sin \theta}{r^{2}}\)
    But θ = 90
    °
    or, \(dB = \frac{\mu_{o}}{4 \pi} \frac{i dl }{r^{2}}\) ............(i)
    The induced magnetice field is perpendicular to the current element and radius vector(r) which can be resolved into two components,
    ​​​​​​For every current element dl there exist an diametrically opposite current element dl' that induces dB' magnetic field at point P which can be resolved into two components,
    • dB' cos φ along vertical
    • dB' sin φ along horizontal , as shown in the figure.
    The vertical component of induced magnetic field cancel each other and resultant magnetic field is given by the horizontal component of induced magnetic field.
    i.e (dB)R = dB sin φ ............(ii)
    From equation (i) and (ii),
    \((dB)_{R} = \frac{\mu_{o}}{4 \pi} \frac{i dl }{r^{2}} \sin \phi\) ............(iii)
    From figure,
    \(\sin \phi = \frac{a}{r}\) ............(iv)
    From equation (iii) and (iv),
    \((dB)_{R} = \frac{\mu_{o}}{4 \pi} \frac{i dl }{r^{2}} \frac{a}{r}\)
    \((dB)_{R} = \frac{\mu_{o}}{4 \pi} \frac{i dl a}{r^{3}} \) ............(v)
    Again from figure,
    r2 = a2 + x2
    i.e. r = (a2 + x2)½
    or, r3 = (a2 + x2)3/2 ............(vi)
    From equation (v) and (vi),
    \((dB)_{R} = \frac{\mu_{o}}{4 \pi} \frac{i dl a}{(x^{2}+a^{2})^{\frac{3}{2}}} \) But such current elements vary from 0 to 2 π a,
    \(Induced\ Magnetic\ Field(B) = \int_{0}^{2 \pi a}dB\)
    \(= \frac{\mu_{o} i a}{4 \pi (x^{2}+a^{2})^{\frac{3}{2}}} \int_{0}^{2 \pi a} dl \)
    \(= \frac{\mu_{o} i a}{4 \pi (x^{2}+a^{2})^{\frac{3}{2}}} [l]_{0}^{2 \pi a} \)
    \(= \frac{\mu_{o} i a}{4 \pi (x^{2}+a^{2})^{\frac{3}{2}}} \times 2 \pi a\)
    \(= \frac{\mu_{o} i a^{2}}{2 (x^{2}+a^{2})^{\frac{3}{2}}} \)
    If coil consist of N no. of turns,
    \( B = \frac{\mu_{o} i N a^{2}}{2 (x^{2}+a^{2})^{\frac{3}{2}}} \)
    which is the required expression.

    Direction : As given by right hand curl rule, the direction of induced magnetic field is towards the axis of the coil.
  3. To calculate the magnetic field induced at a point due to infinitely long straight current carrying conductor
    Let us consider an infinitely long straight conductor carrying current 'i'. We are intrested to calculate the magnetic field induced at point P at the distance of 'x' from the conductor.
    Let us take a small current element 'dy' which is 'r' distance away from the point P i.e \(\overrightarrow{r}\) is a radius vector.
    According to Biot Savart's Law,
    \(dB = \frac{\mu_{o}}{4 \pi} \frac{i dy \sin \theta}{r^{2}}\) ............(i)
    From figure,
    θ + φ = 90
    °
    i.e θ = 90 - φ
    Sin θ = Sin (90 - φ)
    i.e Sin θ = Cos φ ............(ii)
    From equation (i) and (ii),
    \(dB \= \frac{\mu_{o}}{4 \pi} \frac{i dy \cos \phi}{r^{2}}\) ............(iii)
    Again from figure,
    \(\tan \phi = \frac{y}{x}\)
    ⇒ y = x tan φ
    Diff. both sides with respect to φ,
    dy = x Sec2 φ dφ ............(iv)
    From equation (iii) and (iv) ,
    \(dB \= \frac{\mu_{o}}{4 \pi} \frac{i x \sec^{2} \phi d \phi \cos \phi}{r^{2}}\) ............(v)
    Also,
    \(\cos \phi = \frac{x}{r}\)
    ⇒ r = x sec φ
    ⇒ r2 = x2 sec2 φ ............(vi)
    From equation (v) and (vi),
    \(dB = \frac{\mu_{o} i}{4 \pi x} \cos \phi d \phi\)
    \(B = \int_{\phi_{1}}^{\phi_{2}} dB\)
    When y = -
    , φ1 = - π/2
    When y = 
    , φ2 =  π/2
    \(B = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mu_{o} i}{4 \pi x} \cos \phi d \phi\)
    \(B = \frac{\mu_{o} i}{4 \pi x}  \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \phi d \phi\)
    \(B = \frac{\mu_{o} i}{4 \pi x}  [\sin \phi]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \)
    \(B = = \frac{\mu_{o} i}{4 \pi x}  [\sin \frac{\pi}{2} - \sin (- \frac{\pi}{2})] \)
    \(B = = \frac{\mu_{o} i}{4 \pi x} \times 2 \sin \frac{\pi}{2} \)
    \(B = = \frac{\mu_{o} i}{2 \pi x} \)
    Direction : The magnetic lines of force are circular and concentric to the conductor. A tangent drawn at any point on these lines of force gives the direction of induced magnetic field.
  4. To calculate the magnetic field induced at a center on the axis of current carrying solenoid
    A solenoid is the structure formed by close and compact rings of a conducting wire.
    Let us consider a infinitely long solenoid of radius 'a' having 'n' no. of turns per unit length carrying current 'i' in anti clockwise direction. Let us denote each coils of solenoid by a dot(.)
    We are intrested to calculate the magnetic field induced at point P on the axis of current carrying solenoid.
    Let dB be the small magnetic field induced at point P due to small current element dx,
    Then , \(dB = \frac{\mu_{o} N i a^{2}}{2 r^{3}}\) ............(i)
    The No. of turns in segment dx(N) = n dx
    Now eqn(i) becomes,
    \(dB = \frac{\mu_{o} n dx i a^{2}}{2 r^{3}}\) ............(ii)
    From figure, \(tan \phi = \frac{a}{x_{o} - x}\)
    ⇒xo - x = a cot φ
    Diff. both side with respect to φ
    - dx = - a cosec2 φ dφ
    i.e  dx =  a cosec2 φ dφ ............(iii)
    From equation (iii) and (ii),
    \(dB = \frac{\mu_{o} n  i a^{2} a cosec2 \phi d \phi}{2 r^{3}}\) ............(iv)
    From figure,
    \(\sin \phi = \frac{a}{r}\)
    ⇒r = a cosec φ
    ⇒r3 = a3 cosec3 φ ............(v)
    From equation (iv) and (v),
    \(dB = \frac{\mu_{o} n  i a^{3}  cosec2 \phi d \phi}{2 a^{3} \cosec^{3} \phi}\)
    \(dB = \frac{\mu_{o} n  i}{2} \sin \phi d \phi\)
    \(B = \int_{\phi_{1}}^{\phi_{2}} dB\)
    For infinitely long solenoid, \phi1 = 0
    ° \phi2 = 180°;
    \(B = \frac{\mu_{o} n  i}{2} \int_{0}^{\pi} \sin \phi d \phi\)
    \(= \frac{\mu_{o} n  i}{2} [(- \cos \phi)]_{0}^{\pi}\)
    \(= - \frac{\mu_{o} n  i}{2} (\cos 180 - \cos 0)\)
    \(= - \frac{\mu_{o} n  i}{2} (-1-1)\)
    \(= \frac{\mu_{o} n  i}{2} \times 2\)
    \(B = \mu_{o} n  i\)
    which is the required expression.
    Direction : According to right hand curl rule , direction of induced magnetic field is towards the axis of solenoids.

Ampere's Law
Ampere's Law states that, "The line integral of induced magnetic field over a closed path is μo times the current enclosed by a closed loop."
Mathematically,
\(\oint \overrightarrow{B} \overrightarrow{dl} = \mu_{o} i\)
Proof
Let us consider a long straight conductor carrying current 'i' , that induces B magnetic field at point P at distance x from it.
Then from Biot Savart's Law,
\(B = \frac{\mu_{o} i}{2 \pi x}\) ............(i)
Let us construct a close circle passing through point P and concentric to the current carrying conductor and take an elementary length 'dl' on it's circumference.
Then,
\(\oint \overrightarrow{B}. \overrightarrow{dl} = \oint B\ dl \cos \theta\)
\(= \oint B\ dl\) [Since, θ = 0
°]
\(= B \oint  dl\)
But such elementary lengths vary from 0 to 2
π x.
\(\therefore \oint \overrightarrow{B}.\overrightarrow{dl} = B\ [l]_{0}^{2 \pi x}\)
\( \oint \overrightarrow{B}.\overrightarrow{dl} = B\ 2 \pi x\) ............(ii)
From equation (i) and (ii) we get,
\( \oint \overrightarrow{B}.\overrightarrow{dl} = \frac{\mu_{o}i}{2 \pi x}. 2 \pi x\)
i.e. \(\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{o}i\)
which is Ampere's Law.

Application of Ampere's Law

  1. To calculate the magnetic field induced at a point due to long straight current carrying conductor
    Let us consider a long straight conductor carrying current 'i'. We have to calculate the magnetic field induced at point P at a distance 'x' from the conductor.
    To apply Ampere's Law, let us construct a close circle passing through point P and concentric to the current carrying conductor. Let, dl be the elementary piece on the circumference of the circle as shown in the figure below.
    Then, From Ampere's Law,
    \(\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{o}i\)
    \(\oint B.dl \cos \theta = \mu_{o}i\)
    But, θ = 90
    °
    \(\oint B.dl = \mu_{o}i\)
    \(B \oint dl  = \mu_{o}i\)
    Such elementary piece varies from 0 to 2
    π x.
    \(\therefore B\ [l]_{0}^{2 \pi x} = \mu_{o}i\)
    \(B . 2 \pi x = \mu_{o}i\)
    i.e \(B = \frac{\mu_{o}i}{2 \pi x}\) ............(i)
    which is the required expression.

  2. To calculate the magnetic field induced at a point on the axis of long current carrying solenoid
    A solenoid is a coil wound closely to form a helix. Let us consider s long solenoid having 'n' number of turns per unit length carrying current 'i' in anti clockwise direction.
    Let us denote coild of solenoid by the dots.
    We have to calculate the magnetic field induced at point P on the axis of solenoid. To apply Ampere's Law, let us construct a rectangle of length x that passes through point P as shown in the figure.
    Now,
    No. of turns enclosed by the rectangle = n x
    Current enclosed by the rectangle (i') = i n x
    From Ampere's Law,
    \(\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{o}i\) ............(i)
    Here, \(\oint \overrightarrow{B}.\overrightarrow{dl}=\oint_{W}^{X}\overrightarrow{B}.\overrightarrow{dl} + \oint_{X}^{Y}\overrightarrow{B}.\overrightarrow{dl} + \oint_{Y}^{Z}\overrightarrow{B}.\overrightarrow{dl} +\oint_{Z}^{X}\overrightarrow{B}.\overrightarrow{dl}\) ............(ii)
    Now, \(\oint_{W}^{X}\overrightarrow{B}.\overrightarrow{dl} = \oint_{W}^{X} B\ dl \cos \theta\)
    \(= \oint_{W}^{X} B\ dl \cos 0 \degree \)
    \(= \oint_{W}^{X} B\ dl \)
    \(B [l]_{0}^{x}\)
    =Bx
    Further, \(\oint_{X}^{Y}\overrightarrow{B}.\overrightarrow{dl} = \oint_{X}^{Y} B\ dl \cos \theta\)
    \(\oint_{X}^{Y} B\ dl \cos 90 \degree\) = 0
    \(\oint_{Y}^{Z}\overrightarrow{B}.\overrightarrow{dl} = 0 \)[Magnetic field outside the solenoid is zero.]
    \(\oint_{Z}^{W}\overrightarrow{B}.\overrightarrow{dl} = \oint_{Z}^{W} B\ dl \cos \theta\)
    \(\oint_{Z}^{W} B\ dl \cos 90 \degree\) = 0
    Now equation (ii) becomes,
    \(\oint\overrightarrow{B}.\overrightarrow{dl} = Bx + 0 + 0 + 0  \)
    i.e. \(\oint\overrightarrow{B}.\overrightarrow{dl} = Bx \) ............(iii)
    From equation (i) and (iii) ,
    B x =
    μo i n x
    B =
    μo n i 
    which is the required expression.