Ampere's Circuital Law 5t446n

Class XII/PHYSICS

1. Magnetic field at the centre of a circular loop Consider a circular loop of radius a, centre O and carrying current I as shown. To find the magnetic field at O, consider an infinitesimally small length AB=dl of the loop. the centre O to the middle point C of the current element AB. ∠ ACO=90°. According to Biot-Savart’s law, the magnetic field at O due to AB is

  µ0 Id l ×rˆ , where rˆ is the unit vector along CO dB = 4π a2  Since the angle between Id l and rˆ is 90°, the magnitude of magnetic field at point O due to AB is given by-

µ 0 Idl sin 90° µ 0 Idl = 4π 4π a 2 a2

dB =

since, sin 90° = 1

Hence the magnetic field at O due to the whole circular loop, is given by-

B = ∫ dB = ∫ Now,

∫ dl

∫ dl

= 2πa , the circumference of the circular loop.

Therefore, B = Or, B =

µ 0 Idl µ 0 I = 4π a 2 4π a 2

µ0 I 2 πa 4π a 2

µ 0 2 πI 4π a

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

In case the circular loop has n turns, then- B =

µ0 2 πn I 4π a

------------------ (2)

If the current flows in the clockwise direction, then the magnetic field at its centre is perpendicular to the plane of the loop and in inward direction. Else, if the current flows in the anti-clockwise direction, then the magnetic field at its centre is perpendicular to the plane of the loop and in outward direction. AMPERE’S CIRCUITAL LAW The statement of Ampere’s circuital law is:  “The line integral of magnetic field (induction) , B around any closed path in a magnetic field is equal to µ0 (magnetic permeability) times the current threading through the loop.”   Mathematically, ∫ B • dl = µ0 I ----------------------- (3) Applications of Ampere’s circuital law 1. To find magnetic field due to a very long and straight conductor carrying electric current Consider a very long (in principle infinitely long) straight conductor carrying electric current I as shown in the figure. Suppose the point P is at a distance r from the conductor. Draw a circle, ing through P.   Line integral of the magnetic field along the circular path= ∫ B • dl   According to Ampere’s circuital law, ∫ B • dl = µ0 I -------- (4)





The angle between B and dl is zero. Therefore,   ∫ B • dl = ∫ Bdl cos 0° = ∫ Bdl = B ∫ dl

∫dl

=2πr , the circumference of the circular path of radius r.   Therefore, ∫ B • dl =B.2πr -------------------- (5)

But,

From equations (4) and (5), we haveB.2πr = µ0 I ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------SRHSS/2010/ MAGNETIC EFFECTS OF ELECTRIC CURRENT Page 1 of 5 Prepared by: Sandhya.K

Class XII/PHYSICS

µ I µ 2I B = 0 = 0 . ----------------- (6) 2πr 4π r

Or,

It gives magnetic field due to a very long and straight conductor carrying electric current at a point distant r from it. 2. Magnetic field due to a long straight solenoid

A helical coil consisting of closely wound turns of insulated conducting wire is called a solenoid. Consider a long straight solenoid having n turns per unit length and wound on a cylindrical former. When a current is ed through a solenoid the magnetic field is produced, which is strong and uniform inside, while it is negligibly weak outside. At well inside the solenoid, the magnetic field is uniform and parallel to the length of the solenoid.  To calculate B consider a rectangular path A, B, C and D. let the length of the path be AB=L. Number of turns of the solenoid enclosed by the rectangle is equal to nL. The total current threading through the rectangular path is nLI. According to Ampere’s circuital law, line integral of magnetic field along the rectangular path ABCD is equal to µ0 ( nLI ) i.e.





∫ B • dl

ABCD

= µ0 ( nLI ) --------------- (7)

 D   A     B   C  B • d l = B • d l + B • d l + ∫ B • dl +∫ B • dl -------------- (8) ∫ ∫ ∫

Now,

ABCD

A

B

C

D

The magnetic field on part CD of the loop will be zero as it is lying outside the solenoid. Also some part of AD and BC is outside and the part inside is perpendicular to the magnetic field. Hence magnetic field on them is zero.

  C   A   ∫ B • dl = ∫ B • dl = ∫ B • dl = 0 D

C

B

D

 B    B • d l = B • d l ∫ ∫

Therefore, eqn. (8) becomes,

ABCD

A









Along the path AB, B and dl are along same direction, the angle between B and dl is zero. Therefore, B B  B  B • d l = Bdl cos 0 ° = Bdl = B ∫ ∫ ∫ ∫ dl B

A

A

A

A

B

Now,

∫ dl

= L , the length of the side AB of the rectangular path ABCD. Hence,

A





∫ B • dl

= BL

ABCD

Therefore, equation (7) becomes, BL = µ0 ( nLI ) Or,

B = µ0 nI --------------------- (9)

At points near the ends of the solenoid, the magnetic field is given by-

B=

1 µ0 nI 2

-------------------- (10)

FORCE ON A CHARGE MOVING IN UNIFORM MAGNETIC FIELD ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------SRHSS/2010/ MAGNETIC EFFECTS OF ELECTRIC CURRENT Page 2 of 5 Prepared by: Sandhya.K

Class XII/PHYSICS

When a charged particles moves across a magnetic field, a magnetic force is experienced by the charged particle. The magnitude of force on the charge depends on the following factors: 

Magnetic force is directly proportional to the magnitude of the charge,



Magnetic force is directly proportional to the intensity of magnetic field or magnetic flux density.



Magnetic force is directly proportional to the component of velocity along a direction perpendicular to the direction of the magnetic field,

Fα

q

Fα B

F α vSinθ

Combining above statements:

F α BqVSinθ F = k BqVSinθ

Or,

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

Where k is the constant of proportionality. In S.I when B is measured in tesla, q in coulomb and v in metre second-1, the value of k is found to be 1. Therefore, F = BqVSinθ ------------- (2) The force on a charged particle moving in a magnetic field is called magnetic Lorentz force. Equation (2) gives the magnitude of magnetic Lorentz force. In vector notation,

(

)

   F = q v × B ------------------- (3) Thus, if the charged particle moves along X-axis with a velocity v and the magnetic field is directed   along Y-axis, then the force on the charge +q acts in the direction of v × B , that is, along Z-axis.

(

)

SPECIAL CASES:  When the charged particle is at rest: If the particle is at rest, then v=0. From equation (2), the force on the charged particle, F = Bq (0) sin θ = 0 Thus, inside a magnetic field, a stationary charged particle does not experience any force due to the magnetic field. When the charged particle moves parallel to the magnetic field: If v is parallel to B then, θ=0° or 180°. Then, From equation (2), the force on the charged particle, F = Bqv (0) = 0 [Since sin0°=sin180°=0] Thus, inside a magnetic field, a charged particle does not experience any force due to the magnetic field, if it moves parallel to the magnetic field. 



When the charged particle moves perpendicular to the magnetic field: If v is perpendicular to B then, θ=90°. In that case, From equation (2), the force on the charged particle, F = Bqv (sin 90 °) = Bqv .1 = Bqv (Maximum)

Unit of strength of magnetic field: In S.I, the unit of strength of magnetic field is tesla (T). From equation (2), F B= qv sin θ Therefore,

1tesla =

1N C =1NA −1 m −1 (Since = A ) 1C ×1m / s ×1 s

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Class XII/PHYSICS

The strength of the magnetic field is said to be one tesla, if a charge of one coulomb, when moving with a velocity of 1m/s along a direction perpendicular to the direction of the magnetic field, experiences a force of one Newton. 1 tesla (T) = 1 weber metre-2 (Wb m-2) 1 Gauss (G) = 10-4 tesla (T) LORENTZ FORCE   If an electric field E is also present along with B , the resultant force acting on the charged particle will be,     F = qE + q v × B ---------------------- (4) The force given by above equation is called Lorentz force on the charge q. [The total force experienced by a charged particle in a combined electric and magnetic field, is called the Lorentz force on the charged particle.]

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)

MOTION OF A CHARGED PARTICLE INSIDE A UNIFORM MAGNETIC FIELD.

When a charged particles moves across a magnetic field, a magnetic force is experienced by the charged particle is given by   F =q v ×B Motion of the charged particle can be studied under two cases:   (a) When v is perpendicular to B : Consider a uniform magnetic field along Y-axis. Let the   particle move in the magnetic field such that v is perpendicular to B . The force acts on the particle perpendicular to its velocity, the force performs no work. Hence, the magnitude of the velocity remains constant and the force acting on the particle only changes the  direction of its velocity. Hence, F acts as a centripetal force and makes the charged particle move along a circular path. If r is the radius of the circular path, thenMagnetic Lorentz force= centripetal force

(

(

)

)

2   q v × B = mv ------------------ (5)



r



Since, v is perpendicular to B , we have  q (v ×B ) =qvB sin 90 °=qvB -------- (6) Using eqn. (6) in eqn. (5),

qvB =

mv 2 r

mv ----------------------- (7) qB (7), it is clear thatLarger v, larger the radius of the circular path, and Heavier the particle (large m), larger the radius of the circular path. Greater the strength of the magnetic field (B large), smaller the radius of the circular path.

r= From eqn.   

Period of the circular motion of the charged particle is given bymv 2π qB dis tan ce 2πr 2πm --------------------- (8) T = = = = speed v v qB The angular frequency of the charged particle moving inside the uniform magnetic field is given by-

ω=

qB 2π 2π = = 2πm T m --------------------------- (9) qB

The angular frequency, ω, is called cyclotron frequency.

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Class XII/PHYSICS

From equations (8) and (9), the period of the circular motion and the angular frequency of a charged particle moving inside the uniform magnetic field neither depends upon the magnitude of its velocity nor upon the radius of the circular path.

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