The magnetic field at a point is directly proportional to the current, element length, and sine of the angle, and inversely proportional to the square of the distance.
The linear distance travelled by the charge particle in one revolution or in one time period along the external magnetic field direction is called 'pitch of helix'.
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Magnetic Effects Of Current and Magnetism
The magnetic effect of current is the phenomenon where an electric current creates a magnetic field around a conductor. This was first observed by Hans Christian Oersted. Laws like Biot-Savart and Ampere’s Circuital Law help us understand how these fields are formed. Magnetic field lines show the direction of the field around the wire. Devices like solenoids and toroids use this effect to generate uniform magnetic fields. Depending on how materials respond to these fields, they are classified as ferromagnetic, paramagnetic, or diamagnetic.
Related Video:
1.0Concept of Field and Oersted Experiment
A region around any physical quantity where another similar physical quantity experiences force or torque is called a field.
2.0Oersted Experiment
When the direction of current in the conductor is reversed, then the deflection of the magnetic needle is also reversed. Increasing the current or moving the needle closer increases its deflection.
3.0Biot Savart Law
The magnetic field at a point is directly proportional to the current, element length, and the a sine of the angle, and inversely proportional to the square of the distance.
dB=4πμ0r2IdlSinθ
μ0=4π×10−7Tma−1
μ0: Permeability of free space
dB=4πμ0r3I(dl×r)⇒Vector Form
B=∫dB=4πμ0∫r3I(dl×r)(Integral Form)
4.0Application of Biot-Savart Law
Magnetic Field due to Thin Wire of Finite Length
B=4πaμ0I[sinϕ1+Sinϕ2]inwards
Magnetic Field due to Infinite Straight Wire
B=2πaμ0I
Magnetic Field due to Semi Infinite Straight Wire
Case-1
ϕ1→90oorϕ2→0°
B=4πaμ0I−⊗
Case-2.
ϕ1→90oorϕ2→θ
B=4πaμ0I(1+sinθ)−⊗
Case-3.
ϕ1→90oorϕ2→−θ
B=4πaμ0I(1−sinθ)−⊗
Case-4.
ϕ1→−θorϕ2→900
B=4πaμ0I(1−sinθ)−⊗
5.0Magnetic Field due to Circular Loop
Direction: Can be obtained by right hand thumb rule -Curl the right-hand fingers along the current; the thumb shows the magnetic field direction.
Magnetic Field due to Circular Coil
For N loop- BO=2Rμ0NI
Magnetic Field due to Circular Arc
B=4πμ0RIα
Wire of circuit loop is uniform
BadcBabc=(l1l2)(l2l1)=11
The magnetic field produced by both the arc is at the centre of the circuit loop is equal in intensity and in reverse direction. So (BO)net=0 (always and it is free from angle of connection of terminals)
Wire of circuit loop is non-uniform
Case-1. Thickness-Same, Material - different
Case-2. Thickness-Different, Material - Same
(BO)net=4rμO(I1−I2)
(BO)net=4rμO(I1−I2)
Note: When current divides in any symmetrical planar loop made with uniform wire, then the magnetic field at the centre due to this loop is ZERO.
Magnetic Field at the Axis of Circular loop
BP=2(R2+x2)23μ0IR2=(1+R2x2)23Bo
Case-1: At very large distance from centre, BP=2x3μOIR2
Case-2: Near centre of the ring BP=(1+R2x2)23Bo
6.0Ampere’s Circuital Law and it's Utilities
Ampere’s Circuital Law: Line integral of magnetic field along any closed loop is equal to o times the net current crossing the surface bounded by the loop.
∮B.dl=μ0Ienc
Note:
1. It is applicable only for a steady/constant current
2. It can be applied for any distribution of current but, it is applied for symmetric distribution for calculation purposes.
Magnetic Field due to Long Thin wire
B=2πrμ0I
Magnetic Field due to Long Thick wire
BOut=2πrμ0I
Bsur=2πRμ0I
Bin=2πR2μ0I=2μ0Jr
The magnetic field is maximum at the surface of the wire.
Magnetic Field due to Long Hollow Cylindrical Conductor
At point A, (r<a)
Ienc=0⇒Bin=0
At point B, (r≥b)
Ienc=I⇒Bout=2πrμ0I
At point C, (b>r≥a)
Bmid=2πrμ0I(b2−a2r2−a2)
7.0Magnetic field due to Solenoid
It is a wire wound into a helix with insulated turns. The magnetic field inside is strong and aligned along the axis, while outside it is nearly zero. The field direction inside is determined using the right-hand thumb rule.
Magnetic Field Inside a Long Solenoid
B=μOnI=μO(lN)I
M.F. outside the solenoid B=0
M.F. at the edges/end points B=2μOnI
Key Note: A solenoid creates a magnetic field similar to that of a bar magnet, acting as a magnetic dipole and serving as an electromagnet in various devices.
8.0Magnetic field due to Toroid
A toroid can be considered as a ring-shaped closed solenoid also called endless solenoid. Toroid is a solenoid bent in ring shape.
Point A: Magnetic field in the empty space surrounded by toroid (r<R1),B=0
Point B: Magnetic field outside the toroid (r>R2),B=0
Point C: Inside the Toroid (R1≤r≤R2)B=μOnI
n=2πrmN[rm=2R1+R2=Mean Radius of Toroid]
9.0Motion of Charge in a Magnetic Field
Magnetic Force on Moving Charge
F=q(v×B)=qvBsinθ
Magnetic force depends on angle between v and B
Case-1: θ = 0° or 180°
Case-2: θ = 90°
F=0
Fmax=qvB
Note: Direction of force can be identified by right hand thumb rule or right hand palm rule.
Motion of a Charged Particle in an Unvarying Magnetic Field
Case 1
θ = 0°/180°
Straight Line Path
Case 2
θ = 90°
Circular Path
Case 3
θ ≠ 0°, 90°, 180°
Helical Path
Case 1:
θ=0°
F=qvBsin0°=0
Straight Line Path
θ=180°
F=qvBsin1850°=0
Straight Line Path
Case 2: Motion of charge particle in uniform transverse magnetic field
Radius of Curvature,R=qBmv(∵p=mv)
R=qBp=qB2mKK:Kinetic Energy
K.E, K=qVacc
R=qB2mqVacc⇒R=B1q2mVacc
Time period (T):T=qB2πm
Frequency,f=T1=2πmqB
Angular frequency,ω=mqB
Case-3.Motion of Charge Particle in Oblique Magnetic Field ( 0°,90°,180°)
Radius of Circular Path, R=qBmvsinθ
Time period of circular motion,T=qB2πm
Pitch of helix (p): The linear distance travelled by the charge particle in one revolution or in one time period along the external magnetic field direction is called 'pitch of helix'. T=qB2mvcosθ
Lorentz Force: When a charge moves in an electric field (E) anda magnetic field (B), both forces act on it. The combined effect is called the Lorentz force.
Fnet=q(v×B)+(qE)
10.0Magnetic Force on a Current Carrying Wire
dF=I(dl×B)
F=I(Leff×B)Leff is the displacement vector from starting point of current to end
point of current.
11.0Magnetic Interaction Between Two Parallel Current Carrying Wires
Magnetic force per unit length of each conductor is, dldF12=dldF21=2πdμ0I1I2
12.0Definition of Ampere
Magnetic force/unit length for both infinite length conductor gives as
f=2πdμ0I1I2=2π(1)(4π×10−7)((1)1)=2×10−7N/m
'Ampere is the current which, when passed through each of two parallel infinite long straight conductors placed in free space at a distance of 1 m from each other, produces between them a force of 2 × 10–7 N/m
13.0Bar Magnet and Magnetic Dipole
Bar Magnet: It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. Two poles are present in a Bar magnet, North Pole and South Pole.
Magnetic Dipole: A magnetic dipole consists of a duo of magnetic poles of equal and opposite strength separated by a small distance.
Example: Magnetic needle, Bar Magnet, Current carrying coil/solenoid etc.
M=ml
Vector quantity
Direction: South → North
Unit: A−m2 or J/T
Magnetic Moment of Current Carrying Coil (Loop)
Magnetic Moment M=NIA
14.0Torque and Force on Magnetic Dipole
Bar Magnet
τ=M×B=MBsinθ
τ=MBsinθ
Case-1. Ifθ=90°⇒τ=MB(maximum)
Case-2.Ifθ=0°or180°⇒τ=0(minimum)
Coil or Loop
τ=NIABsinθ
Case-1. Ifθ=90°⇒τ=NIAB(maximum)
Case-2.Ifθ=0°or180°⇒τ=0(minimum)
15.0Moving Coil Galvanometer
The galvanometer has a coil with many turns, free to rotate in a uniform radial magnetic field. A soft iron core strengthens and radializes the field, while a spiral spring resists the coil's rotation.
A current-bearinga coil in a magnetic field experiences a torque.τ=NIABsinθ
The spring S provides a counter torque C that balances the magnetic torque, τ′=Cϕ
In equilibrium,Cϕ=NIAB⇒I=NABCϕ, I∝ϕ
It means the deflection produced is proportional to the current flowing through the galvanometer.
Current Sensitivity
The deflection produced in the galvanometer when a unit current flows through
Voltage Sensitivity
The deflection per unit voltage applied across the voltmeter.
C.S.=Iϕ=CNABrad/A
V.S.=Vϕ=IRϕ=CRNAB(rad/V)
16.0Potential Energy of Magnetic Dipole
Work Done in Rotating the Coil in a Uniform Magnetic Field
W=MB(cosθ1−cosθ2)
Potential Energy of the Coil in Uniform Magnetic Field
U=−MBcosθ
U=−M.B
Case-1
Case-2
Case-3
When MandB are parallel
θ= 0°
τ = 0
Umin=−MB
When MandB are perpendicular
θ=90°
τ=MB
U = 0
When MandB are anti- parallel
θ=180°
τ=0
Umax=MB
17.0Atomic Magnetism
Effective current
Induced Magnetic Field at center
Magnetic Moment
I=2πrqv
Bin=4πr2μoqv
M=2qvr
18.0Angular SHM of Magnetic Dipole
T=2πMBI I:Moment of Inertia =12Ml2
19.0Terminology Used In Magnetism
Magnetizing field or Magnetic Intensity(H): Field in which a material is placed for magnetization.
H=μOB0=A/m
Intensity of magnetization (I): The intensity of magnetisation is the induced dipole moment per unit volume when a magnetic material is placed in a magnetising field.
I=VM⇒A/m
Magnetic susceptibility (χm): It is a scalar with no units & dimensions. Physically it represents the ease with which a magnetic material can be magnetised.
Q-4. Two symmetrical current carrying rings are placed perpendicular to each other with a common centre. If magnetic field at the centre due to one coil is B, then find a net magnetic field.
Solution:
BH=2Rμ0I=Bj^
BV=2Rμ0I=Bi^
Bnet=BH2+BV2⇒B2
Q-5.A hollow cylindrical wire carries a current 5A, having inner and outer radii 'R' and 2R respectively. The magnetic field at a point which is 3R/2 distance away from its axis is (R = 5m).
