CBSE Class 12 Physics Formulae Magnetic Effect Of Current And Magnetism Notes

CBSE Class 12 Physics Formulae Magnetic effect of current and magnetism. Learning the important concepts is very important for every student to get better marks in examinations. The concepts should be clear which will help in faster learning. The attached concepts will help the student to understand the chapter and score better marks in the examinations. 

3.MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

GIST

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM:

1. Magnetic field:

    It is a region around a magnet or current carrying conductor in which its magnetic influence can be felt by a magnetic needle.

2. Biot‐Savart Law

    dB =μ0IdlSinθ/4πr2

    μ0=4π x 10‐7 Tm/A

[Direction of dB can be found by using Maxwell’s Right hand thumb rule.]

3. Applications :

   Magnetic field at a centre of a current carrying circular coil B= μ0I/2a

   Magnetic field at a point on the axis of current carrying coil. B= μ0Nia2/2(a2+x2)3/2 (N=no. of turns in the coil)

4. Ampere’s circuital law

   It states that the line integral of magnetic field around any closed path in vacuum/air is μ0 times the total current threading the closed  path.

  ∫ B. dl= μ0 I

5. Applications

i)  Magnetic field due to straight infinitely long current carrying straight conductor. B= μ0 I/2πr

ii) Magnetic field due to a straight solenoid carrying current B= μ0n I N= no. of turns per unit length

iii) Magnetic field due to toroidal solenoid carrying current.

     B= μ0N I / 2πr

     N= Total no. of turns.

6. Force on a moving charge [ Lorentz Force]

(i) In magnetic field F=q(V x B)

(ii) In magnetic and electric field F=q[E+(ν x B)] Lorentz force

7. Cyclotron

(i) Principle

(a) When a charged particle moves at right angle to a uniform magnetic field it describes circular path.

(b) An ion can acquire sufficiently large energy with a low ac voltage making it to cross the same electric field repeatedly under a strong  magnetic field.

(ii) Cyclotron frequency or magnetic resonance frequency

     ν=qB/2πm, T=2πm/Bq; ω=Bq/m

(iii) Maximum velocity and maximum kinetic energy of charged particle.

     Vm=Bqrm/m 

     Em=B2q2rm2 / 2m

Also, download cbse class 12 physics chapter wise questions with solutions in PDF format.

8. Force on a current carrying conductor in uniform

    F= (I l x B)

    l=length of conductor

Direction of force can be found out using Fleming’s left hand rule.

9. Force per unit length between parallel infinitely long current carrying straight conductors.

    F/l= μ0 I1 I2/2πd

(a) If currents are in same direction the wires will attract each other.

(b) If currents are in opposite directions they will repel each other.

10. 1 Ampere – One ampere is that current, which when flowing through each of the two parallel straight conductors of infinite length and placed in free space at a distance of 1m from each other, produces between them a force of 2x10‐7 N/m of their length.

11. Torque experienced by a current loop in a uniform B.

      τ = NIBA Sinθ

      τ=MXB

     Where M=NIA

12. Motion of a charge in

(a) Perpendicular magnetic field F=q(vxB),F=qvBSin90=qvB (circular path)

(b) Parallel or antiparallel field F=qvBSin0 (or) qvBSin180=0(Straight‐line path)

     If 0<θ<90 , the path is helix

     vCosθ is responsible for linear motion v, v Sinθ is responsible for circular motion

     Hence trajectory is a helical path

13. Moving coil galvanometer It is a sensitive instrument used for detecting small electric Currents.

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

                 I αθ andI = K θ where K= NAB / C

                 Current sensitivity, I s= θ / I=NBA/K

                 voltage sensitivity, Vs= θ /V=NBA/KR

Changing N ‐> Current sensitivity changes but Voltage Sensitivity does not change

(a) Conversion of galvanometer into ammeter

    A small resistance S is connected in parallel to the galvanometer coil

    S=IgG/( I ‐ I g) ; RA=GS/(G+S)

(b) Conversion of galvanometer into a voltmeter.

     A high resistance R is connected in series with the galvanometer coil.

     R=( V/Ig) –G ; Rv=G+R

     Current loop as a magnetic dipole

     Magnetic dipole moment M =evr/2

     M=n( eh / 4πme)

14. Representation of uniform magnetic field.

B

15. Magnetic dipole moment of a magnetic dipole.

The magnetic permeability of a material may be defined as the ration of magnetic induction B to the magnetic intensity H
μ=B/H

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