Magnetic Effect on Currents

Oersted’s Observation: Oersted in I820 observed that a magnetic field is always associated with a current carrying conductor.


Biot Savart Law: Two French Physicists: Biot and Savart, on the basis of experimental observationa, gave a law to find the magnetic field strength due to a current carrying element.

Accordingly the magnetic field strength due to a current carrying element at any point P near it is

(i) Directly proportional to current (i) flowing through it

(ii) Directly proportional to the length of element

Biot Savart Law

Biot Savart Law

(iii) Inversely proportional to the distance of point (P) from the element

(iv) Directly proportional to the sin \theta, where \theta is the angle between current clement and the position vector of point relative to current element

(v) It also depends on medium between the current element and point P.

If we combine all these points, we get magnetic field (\delta B) at point P is:

\delta B \propto = \dfrac{I \delta l sin \theta}{r^2}


Or \delta B = \dfrac{\mu}{4 \pi} \dfrac{I \delta l sin \theta}{r^2}

Where \dfrac{\mu}{4 \pi} is constant of proportionality.

\mu is called absolute permeability of medium.

For free space \mu = \mu = absolute permeability of free space = 4 \pi \times 10^{-7} weber amp^{-1} m^{-1} .

The ratio \dfrac{\mu}{\mu _0} is called the realative permeability of medium and is denoted by \mu r .


\mu _r = \dfrac{\mu}{\mu _0}

In free space the magnetic field induction is :

\delta B = \dfrac{\mu _0}{4 \pi} \dfrac{I \delta l sin \theta}{r^2} \cdots \text{equation 1}


The direction of magnetic field induction \delta \overrightarrow{B} is perpendicular to the plane containing \delta \overrightarrow{l}and\overrightarrow{r} and is given by right hand rule which states “If a conductor carrying current is grasped in the right hand with the thumb pointing in the direction of current, then curling of fingers round the conductor represents the direction of magnetic field“.

In vector form (1) may be expressed as:

\overrightarrow{\delta B} = \dfrac{\mu _0}{4 \pi} \dfrac{I \overrightarrow{\delta l} \times \overrightarrow{r}}{r^3} \cdots \text{equation 2}


For the whole current carrying conductor carrying the resultant magnetic induction:

\overrightarrow{B} = \dfrac{\mu i}{4 \pi} \sigma \dfrac{\overrightarrow{\delta l} \times \overrightarrow{r}}{r^3} Where summation extends for the whole length of conductor.


Unit of Magnetic field


In S.I. system the unit of magnetic field is Tesla or Weber/metre^2 or Newton amp^{-1} meter^{-1}.

The dimensional formula for the magnetic field induction is [ MT^{-2} A^{-1} ]

C.G.S unit of magnetic field is Oersted or Gauss:

1 Tesla = 10^4 Gauss

Sometimes unit of magnetic is used Ampere metrel ^{-1}

1 Tesla = \mu Amp/metre


Magnetic Field Induction in some Special Cases



Magnetic Field Induction

Magnetic Field Induction

(i) Current carrying straight wire: The magnetic field due to straight wire of finite length carrying current ‘I’ is given by:

B = \dfrac{\mu _0 i}{4 \pi R} \cdots \text{equation 1}


Where \alpha and \beta are angles made by upper and lower ends of rod with NP (=R), perpendicular dropped from P on the wire.

If wire is infinitely large \alpha = \beta = \dfrac{\pi}{2}


B = \dfrac{\mu _0 i}{2 \pi R}


The direction of magnetic field is given by right hand rule, according to which. If we spread our right hand in such a way that thumb points along the direction of current and fingers along the point (P), then the perpendicular on palm represents the direction of magnetic field.

Accordingly for a current carrying wire CD of Fig. the field on the right of wire in the plane of paper is downward, while on the left it is upward perpendicular to plane of paper.


(ii) Magnetic field induction at the axis of circular loop of radius ‘a’ a distance x from the center of loop is given by:

B = \dfrac{\mu _0 N I a^2}{2 ( a^2 + x^2 ) ^{\dfrac{3}{2}}}


Where i = current in the loop

N = number of turns in the loop

Magnetic Field of Induction

Magnetic Field of Induction

The direction of field is also the axis of loop. At the center of loop x = 0; therefore, magnetic field at center:

B_{center} = \dfrac{\mu _0 n i}{2a}


If current is anticlockwise, the field is upward perpendicular to plane of loop; which if current is clockwise the magnetic field is downward perpendicular to plane of loop.


(iii) Magnetic field Induction due to a solenoid:

(a) If solenoid has number of tums per meter length n and is of finite length; \alpha and \beta are angles made by lines joining point P from the ends of solenoid with its axis, then magnetic field at P is:

B = \dfrac{ \mu _0 n i}{2} ( cos \alpha - cos \beta )


magnetic Field of Induction

magnetic Field of Induction

(b) For a long solenoid \alpha = 0 , \beta = \pi


B = \mu _0 n I = \dfrac{\mu _0 N i}{l} along the axis.

Here n = (N/l); N = Total number of turns in length l of solenoid.


Force on a moving charge in a uniform magnetic field (Lorentz force)

When a charged particle of charge q moves in a magnetic field of induction B with velocity v, it experiences a magnetic force given by:

\overrightarrow{F} = \overrightarrow{qv} \times \overrightarrow{B} \cdots \text{equation 1}


This is the magnitude of force in qvB sin \theta , \theta being angle between \overrightarrow{v} and \overrightarrow{B} and direction of force is normal to \overrightarrow{v} and \overrightarrow{B} .

This force is also called magnetic Lorentz force.

Case (I): When charged particle moved parallel to field, \theta = \theta


 sin \theta = sin 0 = 0


Or, Force F = 0

That is if charged particle moves parallel to field B, it experience no force and the path of the particle is unperfected straight line.


Case (II): When charged particle (charge q) enters the magnetic field at right angles, \theta = 90^0 , then the force on charged particle is maximum given by:

F=qvB …………………………………………(2)

Circular path

Its direction is normal to \overrightarrow{v} and \overrightarrow{B} , hence no work is done by the magnetic force on the charged particle. Therefore, the kinetic energy and hence speed of charged particle remains constant, but its direction changes. Hence the path of the particle is circular. The necessary centripetal force is provided by the magnetic force.


\dfrac{mv^2}{r} = q v B


Or radius of circular path,

r = \dfrac{mv}{q B}\cdots \text{equation 3}


If T is the period of one revolution, then

T = \dfrac{2 \pi r}{v}


Using (3), we have:

T = \dfrac{2 \pi}{v} [ \dfrac{mv}{qB} = \dfrac{2 \pi m}{q B} ] \cdots \text{equation 5}


And frequency f = \dfrac{1}{T} = \dfrac{q B}{2 \pi m} \cdots \text{equation 6}


Obviously the time period (T) and frequency f are independent of speed v and radius r of circular path.


Case (III): When charged particle enters, the magnetic field in any direction:

If the charged particle enters the magnetic field with velocity \overrightarrow{v} making an angle \theta with magnetic field vector \overrightarrow{B} , then its velocity \overrightarrow{v} may be resolved into two components  v cos \theta and v sin \theta , along and per-pendicular to direction of applied field \overrightarrow{B}.

The component v_{ii} = v \cos \theta is along the field and so remains unaffected by the field and hence the charged particle continues to move along the field with a speed v cos \theta.


The velocity component v \bot = v sin \theta is perpendicular to field and so the path due to this component is circular. The radius of circular path is given by:

r = \dfrac{m v \bot}{q B} = \dfrac{mv sin \theta}{q B} \cdots \text{equation 7}


Thus the resultant motion of the charged particle is due to superposition of two concurrent motions-straight line and circular motions. Due to these motions, the resultant path of the charged particle is a helix (fig b). Time taken by the particle is one revolution.

T = \dfrac{2 \pi r}{v sin \theta} = \dfrac{2 \pi}{v sin \theta} \times \dfrac{m v sin \theta}{q B} = \dfrac{2 \pi m}{q B} \cdots \text{equation 8}



Pitch: The linear distance traversed by the particle along the magnetic field in one revolution is called the pitch of helix and is given by:

P = v cos \theta . T



( v cos \theta ) \times \dfrac{2 \pi m}{q B} = \dfrac{2 \pi m v cos \theta}{q B} \cdots \text{equation 9}





A cyclotron is a device to accelerate charged particles with the help of uniform magnetic field up to energies of the order of million electron-volts (M e V).

It consists of two hollow metallic Dee’s D1 and D2 with a little gap between them. These are placed in a uniform magnetic field (B) perpendicular to the plane of Dee’s. An alternating high voltage source is applied between the Dees.

The charged particle (positive or negative) to be accelerated enters at mid-point C between the Dees. The potential difference applied accelerates the charged particle and the magnetic field keeps the charged particle in circular path.


Condition of resonance: The frequency of rotation of charged particle given by f = \dfrac{q B}{2 \pi m} is made equal to the frequency of alternating voltage source. This is called condition of resonance.

If this condition is achieved, the particle is accelerated twice in a rotation and its speed goes on increasing, hence it continues to move in a circular path of increasing radius and finally it emerges through a hole made in Dee to hit the target.

The energy gained by particle in one rotation =2q V. If N is number of rotations made by the particle in the cyclotron, then total energy gained by particle =N 2q V. if r is radius of Dee, the final velocity v is given by:

\dfrac{mv^2}{r} = q v B = v = \dfrac{q B r}{m}

Therefore, Maximum kinetic energy gained by particle:

E_{max} = \dfrac{1}{2} mv^2 = \dfrac{1}{2} m ( \dfrac{q^1 B^2 r^2}{M^2} ) = \dfrac{q^2 B^2 r^2}{2m}


E_{max} = 2 N q V = \dfrac{q^2 B^2 r^2}{2m}



Force and Torques on currents in a Magnetic Field



(i) Force: Magnetic force on a current carrying conductor of length l, carrying current I, in a magnetic field of strength B is:



F = I l B sin \theta \cdots \text{Equation 1}


\theta being angle between current element \overrightarrow{il} and \overrightarrow{B} .


In vector form,

\overrightarrow{F_m} = \overrightarrow{il} \times \overrightarrow{B} \cdots \text{Equation 2}


Case I: When \theta = 0, magnetic force F=il B sin \theta ^0 = 0

Case II: When \theta = 90^0 , magnetic force  F = il B sin \theta ^0 = il B

This equation give the definition of magnetic field induction B.

Accordingly B = \dfrac{F_m}{il}

If i=1A, l=1m then B = F_m i.e magntic field induction is numerically equal to the magnetic force acting on a conductor of length 1m when it is placed perpendicular to magnetic field and current flowing in it is 1 Ampere.

The direction of magnetic force is normal to current I and field B, given by Fleming left hand rule which states that:

“If forefinger, middle-finger and thumb of the left hand are stretched mutually perpendicular such that forefinger points in the direction of magnetic field, the middle finger along the direction of current, then the thumb points along the direction of force”.

Direction of magnet

Direction of magnet

(ii) Torque: Torque on a rectangle current loop in a uniform magnetic field: if a rectangle loop of Area ( side l, breadth b) carrying current is placed in a uniform magnetic field, then its two sides pq and rs experience no force  and the other two sides ps and rq experience force Bi; each, hence they experience  a couple which tends to rotate the diploe to set its axis parallel to the magnetic field. If \theta is angle made by its normal with the magnetic field,



Perpendicular distance between force (F, F) is  b sin \theta ,


\text{Torque} = Bil ( sin \theta ) = BiA sin \theta


If N is the number of turns in the loop, then:

\text{Torque} = ( N ) B I A sin \theta = B I AB sin \theta


In the vector form \overrightarrow{t} = N I \overrightarrow{A} \times \overrightarrow{B}




Clearly the torque is maximum when sin \theta = 1 or normal to plane of coil makes an angle 90^0 with magnetic field lines i.e., the planes of coil is parallel to the magnetic field lines.


Force between parallel currents



Let a and b be two parallel wires carrying i_1 and i_2 and separated by a distance R. each current carrying wire produces a magnetic field around it and either wire is placed in the magnetic field produced by the other; therefore each wire experiences a magnetic force. The force per unit length on either wire is given by:


Force between parallel current

Force between parallel current

f = \dfrac{F}{\delta l_2} = \dfrac{\mu _0 i_1 i_2}{2 \pi R} \text{Newton/Meter}


If current are parallel, the force is attractive but if currents are anti parallel the force is repulsive.


Definition of Ampere


The force between parallel current carrying wires:

f = \dfrac{\mu _0 i_1 i_2}{2 \pi r} N/m


This expression is used to define a fundamental unit Ampere in S.I. system.

\text{if} i_1 = i_2 = 1 Amp , r = 1 meter \\[3mm] f = \dfrac{\mu _0}{2 \pi} = 2 \times 10^{-7} m

Thus 1 ampere current is defined as the current which when flowing through two parallel thin wires placed at 1 meter apart in vacuum exert a force 2 \times 10^{-7} N on per meter length of either wire.


Moving Coil Galvanometer


A galvanometer is used to detect the current and has moderate resistance.

Principle: When a current carrying coil is placed in a magnetic field, it experiences a torque given by :

t = N I A B sin \theta


Where \theta is the angle between normal to plane of coil and direction of magnetic field. In actual arrangement the coil is suspended between the cylindrical pole pieces of a strong magnet. The cylindrical pole pieces give the field radial such that sin \theta = 1


\text{So torque} = N I A B \cdots \text{equation 1}


Moving coil Galvanometer

Moving coil Galvanometer

If  is torsional rigidly, then for deflection \theta \text{of coil} = C \theta \cdots \text{equation 2}

In equilibrium we have,

External couple= Restoring couple


C \theta = N I A B \text{or} \theta = \dfrac{N A B}{C} i \theta \propto i

In words the deflection produced is directly proportional to current in the coil.

The quality \dfrac{\theta}{i} = \dfrac{N A B}{C} is called the sensitivity of the galvanometer.

For maximum sensitivity number of turns N, coil area A and magnetic field strength B should be higher and torsional rigidity should be low. That is why suspension wire is made of phosphor bronze which has low value of torsional rigidity.




An ammeter is a low resistance galvanometer; used to measure current directly in amperes and is always connected in series with the circuit.



To convert galvanometer into ammeter, a low resistance, called shunt, is connected in parallel to the galvanometer as shown in figure. Let i_g be the current in galvanometer. Let I is the range of ammeter and i_s the current in shunt S. the working equation for conversion of galvanometer into ammeter is:

i_g = \dfrac{S}{S + G} i ( Here i_g < I ) \cdots \text{equation 2}

Value of shunt  S = G \dfrac{i_g}{ ( I - i_g ) }

The resistance of Ammeter R_A so formed is given by :

\dfrac{1}{R_A} = \dfrac{1}{G} + \dfrac{1}{S} or R_A = \dfrac{S G}{S + G} \cdots \text{Equation 3}


Remark: Equation 2 may also be used to increase the range of given ammeter. Then G will be resistance of given ammeter, S shunt applied, i_g its initial range and I the new range desired.




A voltmeter is a high resistance galvanometer and is connected between two points across which potential difference is to be measured i.e. voltmeter is connected in parallel with the circuit.

TO convert a galvanometer into voltmeter, a high resistance R in series is connected with a galvanometer. If V is the range of voltmeter, then:

 i_g = \dfrac{V}{R + G}

Or Value of series resistance:

R = \dfrac{V}{i_g} - G \cdots \text{equation 1}




This is working for conversion of galvanometer into voltmeter. The resistance of voltmeter so formed is:

 R_v = R + G \cdots \text{Equation 2}


Remark: Equation 1 may also be used to increase the range of voltmeter, then:

i_g = \dfrac{V_0}{G} = \dfrac{V}{R + G} \cdots \text{equation 3}


V_0 = \text{initial range} \\ and V = Final range

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