JEE MAIN CHAPTER OVERVIEW
Chapter Priority | 4 (LOWEST) |
Number of problems in PYQ | LOW |
Difficulty | EASY/MODERATE |
To solve JEE MAIN problems for EMI, you need to remember all formulae given in these notes.
Note: Questions can be tricky especially in RC circuit or LR Circuit.
KEY CONCEPTS FOR JEE MAIN
MAGNETIC FLUX
It is a measure of the number of magnetic field lines passing through a surface.
For a uniform magnetic field(B) passing through and area(A) can be defined as:
If the magnetic field has different magnitudes and directions at various parts of a surface:
SI Unit of flux: Weber
METHODS OF CHANGING MAGNETIC FLUX
Change Area of Surface
By increasing or decreasing the surface area, we can change the flux passing through it.
Change Magnetic Field
By increasing or decreasing the magnetic field passing through an area, the flux can be changed.
Change angle between Area and Magnetic Field
We can change magnetic flux by changing the angle between are and magnetic field vector
FARADAY'S LAW
The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
The induced EMF is given by:
In the case of a closely wound coil of N turns, then EMF is given by:
LENZ'S LAW & CONSERVATION OF ENERGY
The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
EXAMPLES:
CASE1: If North-pole of a bar magnet is being pushed towards the closed coil.
As North pole of magnet gets closer, the magnetic flux through the coil increases.
EMF opposes the increase in flux.
Current is in a counter-clockwise direction with respect to an observer situated on the side of the magnet.
CASE 2:
If the North- pole of the magnet is being withdrawn from the coil.
The magnetic flux through the coil will decrease. Induced current in the coil is in clockwise direction and
Coil becomes a magnetic dipole with South-pole faces North-pole of the bar magnet.
This results in an attractive force which opposes the motion of the magnet.
TRY A QUESTION:
Q. A bar magnet is freely falling along the axis of a circular loop as shown in figure. State whether its acceleration a is
less than 'g'
more than 'g'
equal to 'g'
[If you cannot solve any doubts related to any chapter of physics, upload a photo of your doubt at www.savai.co.in or WhatsApp me @ 7982286138, we will send solution.]
JEE MAIN PYQ
Q. If flux is given by: 𝜙 = 8t^2 - 9t + 5
in a closed circuit of resistance 20 𝛺. The magnitude of the induced current at t = 0.25 s will be 'x' mA. Find x.
[Hint: Differentiate flux with respect to time]
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Q. The magnetic flux through a coil perpendicular to its plane is varies according to the relation 𝜙 = (5t^3 + 4t^2 + 2t - 5) Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 sec will be?
[If you cannot solve any doubts related to any chapter of physics, upload a photo of your doubt at www.savai.co.in or WhatsApp me @ 7982286138, we will send solution.]
SOME DIFFERENT VARIETIES OF Qs ASKED IN JEE MAIN
Type 1: Where B is given as a vector in 'i' & 'j' directions.
Q. A conducting circular loop is placed in X - Y plane in presence of magnetic field
If the radius of the loop is 1 m, the induced emf in the loop, at time, t = 2s is n𝜋 V. The value of n is?
[Hint: Dot product of B & A vector needs to be done. What is the direction of A vector? What is the magnitude of A vector? Once you have the dot product, you need to differentiate.]
[If you cannot solve any doubts related to any chapter of physics, upload a photo of your doubt at www.savai.co.in or WhatsApp me @ 7982286138, we will send solution.]
Type 2: Rotating circular loop
Q. A circular coil of 1000 turns each with area 1 m^2 is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07 T. The maximum voltage generation will be __V?
Hint: When a coil with N turns is rotating in a magnetic field with angular frequency 'w' (omega), flux (phi) is given by:
AVERAGE CHARGE TRANSFERRED
EMF(Ɛ) = -∆𝜙/∆t
Current (i) = -∆𝜙/R∆t
∆Q = i∆t
∆Q = -∆𝜙/R
TRY A QUESTION
Q. A square loop ACDE of area 20 cm^2 and resistance 5 Ω is rotated in a magnetic field B = 2T through 180°, in 0.01 s. Find the magnitudes of average values of e, i and ∆q
[Hint: Use the above equations]
MOTIONAL EMF
It is the EMF induced by moving a conductor through a magnetic field.
When a conductor of length L, moves perpendicularly through a magnetic field with velocity v, the EMF is given by:
EMF(Ɛ) = BLv
This equation is only valid when B,L & v are mutually perpendicular
The positive terminal of induced EMF is in the direction of cross product of v & B.
(Right hand thumb rule: Curl the fingers of your right hand from direction of v to B. Direction of thumb gives the direction of cross product and EMF.)
WHEN B,L & v NOT MUTUALLY PERPENDICULAR
We have to take components of vectors, EMF is due to mutually perpendicular components.
EMF in this case is given by:
If L & v are parallel to each other, then EMF induced will be ZERO as shown below.
ROTATION OF CONDUCTOR IN MAGNETIC FIELD
(Very important topic)
1. Conductor of length (L)
2. Rotated with angular velocity 'w' (omega)
3. In a magnetic field (B)
Then, EMF (Ɛ) is given by:
Result is obtained by integration (integration is not relevant for mains)
ENERGY & POWER FOR A CIRCUIT WITH A MOVING CONDUCTOR
Consider a conductor (L) moving in a magnetic field (B) with constant velocity (v).
It is connected to resistor (R) as shown.
A current (I) is induced in the circuit due to motional EMF.
I = BLv/R
In this case, a magnetic force will act on the conductor as it carries a current (I) in a magnetic field. Magnetic Force = FB = IBL
In order to maintain constant velocity of conductor, an external force needs to be applied equal to magnetic force but in opposite direction.
Fext = -FB
Magnitude of FB or Fext is:
Power delivered by this external force = Power delivered to resistor = Heat lost from resistor
Power delivered by this external force = Power delivered to resistor
Heat lost from resistor
TRY A JEE MAIN PYQ
Q. A metallic conductor of length 1m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad/s. If the horizontal component of earth’s magnetic field is 0.2 × 10^(-4) T, then emf induced between the two ends of the conductor is:
[Hint: What is the direction of Earth's magnetic field? Is it perpendicular to the plane of rotation of given conductor? If so, we can use the rotational EMF formula. ]
[If you cannot solve any doubts related to any chapter of physics, upload a photo of your doubt at www.savai.co.in or WhatsApp me @ 7982286138, we will send solution.
SELF INDUCTANCE
When EMF is induced in an isolated coil
- due to a change in the flux through the coil
- by varying the current passing through the same coil
this phenomenon is called self induction
Flux through a coil of N turns is proportional to the current through the coil:
N𝜙 = LI
N = Number of turns in the coil
𝜙 = Flux through coil
I = Current through coil
L = Self Inductance
The induced EMF is given by:
SELF INDUCTACE IN A SOLENOID
Self-inductance of a long solenoid of cross- sectional area A and length l, having n turns per unit length.
If solenoid filled with object of some relative permeability:
ENERGY STORED IN A CAPACITOR
Work Done to build up current in Inductor = Energy Stored in Inductor =
MUTUAL INDUCTANCE
An electric current can be induced in a coil by flux change produced by another coil in its vicinity. This is called mutual inductance.
For example, in the above image changing current in coil 1 can change flux passing through coil 2. This can induce an EMF in coil 2.
Flux in coils 2 due to 1 is proportional to current in coil 1:
N(𝜙2,1) = (M2,1)*I1
M2,1 is called mutual inductance of coil 2 with respect to coil 1.
Always remember,
M2,1 = M1,2
TRY A JEE MAIN PYQ:
Q. A small square loop of wire of side l is placed inside a large square loop of wire L (L >> l). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is?
[Hint: In such questions, (L >> l) is given. It means that you need to find Magnetic field at centre of larger coil and consider this to be constant throughout the area of the smaller coil. Find flux through smaller coil and get it in the form:
𝜙 = MI
where I is the current in larger coil
M is a constant, it is your mutual inductance.]
[If you cannot solve any doubts related to any chapter of physics, upload a photo of your doubt at www.savai.co.in or WhatsApp me @ 7982286138, we will send solution.]
LR CIRCUIT
When an inductance and resistance are joined in series and connected to a DC voltage source.
Current gradually increase to its maximum value.
The current can be described by:
Time constant τ =L/R
[determines the rate at which the current reaches its steady-state value] The variation of current with time is as follows:
Boundary condition in LR circuit:
TIME | CURRENT | KEY POINT |
time = 0 | i = 0 | Inductor acts as a open circuit |
time = infinity | i = V/R | Inductor acts as a short circuit |
RC CIRCUIT
When an capacitor and resistance are joined in series and connected to a DC voltage source.
Variation of charge on capacitor with time in series RC circuit
Time constant τ =RC
Graph of charge versus time is as follows:
Current (I) = dQ/dt
Current in RC circuit varies as:
Initially at t= 0, current is V/R then as time passes current decreases and gradually goes to ZERO at infinite time.
Boundary Cases in RC Circuit
TIME | CURRENT | KEY POINT |
t = 0 | V/R | Capacitor acts as short circuit |
t = infinity | ZERO | Capacitor acts as open circuit |
TRY A JEE MAIN PYQ
Q. In the figure shown, a circuit contains two identical resistors with resistance R = 5Ω and an inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?
[Hint: After a long time inductor acts as a short circuit]
[If you cannot solve any doubts related to any chapter of physics, upload a photo of your doubt at www.savai.co.in or WhatsApp me @ 7982286138, we will send solution.]
Hope you can REVISE EMI & AC FOR JEE MAIN PHYSICS from this guide to answer all varieties of JEE MAIN PYQs. Thanks a lot!
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