Q.A rectangular frame of wire is placed in a uniform magnetic field directed outwards, normal to the paper. AB is connected to a spring which is stretched to A′B′ and then released at time t=0. Explain qualitatively how induced e.m.f. in the coil would vary with time. (Neglect damping of oscillations of spring)
Dnh Dd CbseCBSE Class XII Board 2018Subjective· 2mImportance★★★★★
When a conductor moves through a magnetic field, the free charges inside it experience a magnetic force. This force pushes the charges along the conductor, setting up a potential difference across its ends — a motional emf. It is electromagnetic induction viewed from a moving conductor rather than from a changing field.
Origin: the Lorentz force
Consider a straight rod of length l moving with velocity v perpendicular to a uniform field B. Each free charge q feels a force
F=q(v×B)
of magnitude qvB directed along the rod. Charges pile up at the ends until the electric field they create balances the magnetic force. The resulting potential difference — the motional emf — is
ε=Bvl
Consistency with Faraday's law
Let the rod of length l slide along parallel conducting rails, sweeping out a distance x. The enclosed area is A=lx, so the flux is Φ=Blx. Then
ε=−dtdΦ=−Bldtdx=−Blv
The magnitude Bvl matches the Lorentz-force result, showing that the flux rule and the force picture agree.
Worked example
A rod of length 0.4m moves at 5m/s perpendicular to a field of 0.5T:
ε=Bvl=0.5×5×0.4=1V
If the circuit resistance is 2Ω, the induced current is I=ε/R=0.5A.
Force and energy …
Why this formula?
Motional EMF: Why the Formula Holds
Let's build this from first principles — understanding the why before the formula.
The Core Idea
Motional EMF arises when a conductor moves through a magnetic field. The key insight: moving charges in a magnetic field experience a magnetic force, which acts like a battery pushing charges around the conductor.
Step 1: The Force on a Moving Charge
A charge q moving with velocity v in a magnetic field B feels the Lorentz magnetic force:
Fm=q(v×B)
This force is perpendicular to both velocity and magnetic field.
Step 2: What Happens Inside a Moving Conductor
Consider a straight metal rod of length L moving with constant velocity v perpendicular to a uniform magnetic field B (pointing into the page).
Free electrons in the rod are moving with the rod at velocity v.
Each electron experiences a magnetic force:
Fm=−e(v×B)
(negative sign because electron charge is −e)
This force pushes electrons along the rod — say, toward one end.
Step 3: Charge Separation Creates an Electric Field
As electrons accumulate at one end, that end becomes negatively charged, leaving the other end positively charged.
This charge separation creates an internal electric fieldE inside the rod, pointing from positive to negative end.
The electric field exerts an opposing force on the electrons:
Fe=−eE
Step 4: Equilibrium — The "Battery" is Formed
Charge keeps moving until the electric force balances the magnetic force:
Fe+Fm=0
−eE−e(v×B)=0
E=−(v×B)
Magnitude-wise (for perpendicular v and B):
E=vB
Step 5: From Electric Field to EMF
The motional EMFE is the work done per unit charge to move a test charge from one end to the other:
E=∫negativepositiveE⋅dl
For a uniform field along the rod of length L:
E=E⋅L=vBL
The Key Formula
Motional EMF for a straight conductor moving perpendicular to B:
E=BLv
Why This Makes Physical Sense
Quantity
Role
B
Stronger magnetic field → larger force on charges
L
Longer conductor → more charge separation possible
When released, side AB executes simple harmonic motion, so its displacement (and hence the enclosed area and flux) varies sinusoidally; the induced emf, being −dΦ/dt, is also sinusoidal but 90∘ out of phase with the displacement. …
AB oscillates in SHM ⇒ flux Φ∝cosωt⇒ emf =−dΦ/dt∝sinωt: an undamped sinusoidal emf.
Concept. As the spring-loaded side AB slides in SHM, the area of the frame inside the field changes, changing the magnetic flux linked with the coil. A changing flux induces an emf (Faraday's law).
Why sinusoidal. Take the SHM displacement of AB as x(t)=x0cosωt. If L is the length of AB and B the field, the enclosed area is A(t)=A0+Lx(t), so