Home | Courses | Back to Chapter | Any Question? |

PHYSICS-12

Chapter 14 - Magnetic effect of electric current

Definations:

1. Magnetic field: The space or region around the magnet in which magnetic influence can be experienced.

2. Magnetic field intensity: The magnetic field intensity at a point inside the magnetic is the magnetic force per unit pole strength of north pole.
i.e. B= F/m
where, F= magnetic strength
m= pole strength
Unit of pole strength (m)= Ampere meter (Am)
Unit of B = N/Am
Tesla (T) in SI
Gauss in CGS
[1T = 104 Gauss]
The direction of the magnetic field intensity is away from north and towards south pole.

3. Magnetic lines of force: The imaginary lines which represents the magnetic field. The path followed by north pole of compass needle when it is free to move as shown in figure:



4. Magnetic flux: The magnetic flux associated with the surface is the number of magnetic lines of force passes through that surface.

5. Magnetic flux density: The flux density at a point inside a magnetic field is the magnetic flux passing normally through unit surface area.
i.e. B= /A = weber/m2
Tesla (T)
therefore, = BA, when magnetic lines of force passes normally.



When the magnetic lines of force are not passes normally but makes an angle θ with the normal then magnetic flux is given by:
Φ= AB cosθ

Force on moving charge in uniform magnetic field


Consider a '+q' charge is moving in a uniform magnetic field B, with velocity V, making an angle θ with the direction of magnetic field. Then the charge particle experiences the force known as magnetic lorentz force along z-axis (i.e. perpendicular to the plane of and ). Experimentally it is found that the force is:-
1. Directly proportional to magnetic field.
i.e. Fm ∝ B
2. Fm ∝ q
3. Fm ∝ v
4. Fm ∝ sinθ


Combining all these equations we get,
F m ∝ Bqv sinθ
F m= kBqv sinθ
where 'k' is proportionality constant and its value is found to be 1.
therefore, F m= Bqv sinθ

In vector form,
= q ( x )
i.e. is perpendicular to and
Special Cases:

1. If V=0, Fm=0
i.e. the charge particle at rest doesn't experienced force.

2. If θ= 0o, 180o
F m= 0
i.e. when the velocity of charge particle is parallel or anti-parallel to the direction of magnetic field, then it doesn't experience force and no acceleration is produced in the charged particle and therefore speed of charge particle is uniform and hence kinetic energy remains constant. No work is done by this force in the charge particle. Hence the path followed by the charged particle is straight line.

3. If θ= 90 o, Fm= Bqv (maximum)
When the velocity is perpendicular to the direction of magnetic field, then charge particle experiences maximum force. According to Fleming's left hand rule, F m perpendicular to v, then this force allowed to move the charge particle in a circular path and speed remains uniform and therefore kinetic energy remains constant. No work is done by this force on the charge particle as,
W= Fs cos90 o
therefore, W= 0

4. When the velocity of charge particle is neither parallel, anti-parallel nor perpendicular but makes and angle θ with the direction of magnetic field. Then the path followed by the charge is spiral/helical.

Torque on current carrying rectangular coil in uniform magnetic field.


Consider a rectangular coil PQRS is placed in a uniform magnetic field B. The four sides of the rectangule can be supposed to be four current carrying straight conductor. Here the sides PQ and RS is perpendicular to the direction of magnetic field whereas the sides QR and SP makes an angle θ with the direction of magnetic field.
Let, length of rectangle = PQ = RS = l and
breadth of rectangle = QR = SP = b
The force experiences by the conductor PQ and RS is given by,
F 1= BIl
F 3= BIl
Similarly, the force experiences by the conductor QR and SP is given by,
F 2= BIb sinθ
F 3= BIb sinθ
The direction of this forces is given by fleming's left hand rule.
The forces F2 and F4 are equal and opposite and have same line of action. So, they together cannot constitute a couple to produce torque.
On the other hand, the forces F1 and F 3 are equal and opposite and have different line of action. So, they together can constitute a couple to produce torque in the coil. Therefore the torque due to couple is given by,
𝞃 = Either force x perpendicular distance between the forces = F 1 or F3 XYT



𝞃 = F 1 or F3 x b cosθ
BIl x bcosθ
𝞃 = BIA cosθ
if the coil is N time turns,
𝞃 = BINA cosθ

Special Cases:

1. If θ= 0 o, 𝞃= BINA (maximum)
2. If θ= 90o, 𝞃= 0 (no torque)

Hall effect:


When current is passed through a conductor along x-axis and the conductor is placed in a uniform magnetic field along y-axis then the potential difference p.d. is produced in between the conductor such that electric field is produced in between the conductor along z-axis. This effect is known as Hall-effect. The p.d. produced is call Hall voltage and the field is called Hall field.
Let 'I' be the current through the conductor along x-axis and the conductor having breadth 'b' and thickness 't' is placed in a uniform magnetic field 'B' along y-axis. Let, vd be the drift velocity of the electrons then the magnetic force on each electron is given by,


F m= Bev <---------- (i) (along z-axis in the upward direction)

Due to this force, the electrons inside the conductor tends to move in the upward direction leaving excess +ve charge at the lower surface and excess negative charge at the upper surface. Due to this separation of charges, p.d. is produced inside the conductor. As a result, electric field is produced inside the conductor along z-axis.
The flow of electrons in the upward direction will continue until the magnetic force on the electron must be balances by electric force on the electron (downward direction). Let 'E' be the strength of electric field then electric force on each electron is given by,

F e= eE <------- (ii)
At equilibrium,
F m = Fe
Bev = eE
v= E/B <---------- (iii)
The current flowing through the conductor is given by,
I= venA
I= (E/B)en bt
I= (vH/b.B) enbt
v H= (IB)/(ent)
Now,
current density is given by
J= I/A
J= (venA)/A
J= ven
J= (E/B)en
1/ne = E/JB = R H= Hall coefficient


In case of any problem ask me in qustions section!!!!

© 2020 onlinelearnal.com