Analyzing the Setup
Imagine a long, straight wire carrying a steady current I. This wire acts like a magnetic fountain, creating a magnetic field that permeates the space around it. Now, picture a semi-circular conducting rod placed on two parallel rails, moving upwards with a constant velocity v. As this rod slices through the invisible magnetic field lines, it experiences a phenomenon known as motional EMF.
However, there is a catch here. The magnetic field produced by a straight wire is not uniform; it gets weaker as you move further away. According to Ampere's Law, the magnetic field at a distance r is given by:
Because the field varies along the length of the rod, we cannot simply use the standard formula E=Blv. We must rely on the power of calculus to find the total induced EMF.
The Master Equation for Motional EMF
To find the total EMF, we consider a tiny, infinitesimally small element of the rod of length dr, located at a distance r from the wire. The motional EMF across this tiny element is:
Substituting the expression for the magnetic field, we get:
To find the total EMF generated across the entire rod, we must integrate this expression from the position of the first rail (r1=1 cm) to the second rail (r2=4 cm):
Pulling the constants out of the integral, we are left with the integral of 1/r, which evaluates to the natural logarithm:
E=2πμ0Iv[lnr]14=2πμ0Iv(ln4−ln1)
Since ln1=0 and ln4=2ln2, the expression simplifies beautifully to:
Now, let's substitute the given numerical values into our elegant equation. We know μ0=4π×10−7, I=2 A, v=3 m/s, and ln2=0.7:
E=24×10−7×0.7=1.68×10−6 V
The RC Circuit Dynamics
With the rod moving at a constant velocity, it acts exactly like a constant DC battery with an EMF of 1.68×10−6 V. This "battery" is connected in series with a resistor R=1.4 Ω and an initially uncharged capacitor C0=5.0 μF.
When the motion starts at t=0, the uncharged capacitor offers zero opposition to the flow of charge—it acts like a perfect short circuit. Therefore, the current in the circuit is at its absolute maximum right at the beginning. We can find this maximum current using Ohm's Law:
imax=RE=1.41.68×10−6=1.2×10−6 A
This confirms that Option (A) is correct.
As time progresses, charge accumulates on the capacitor plates. This buildup creates an opposing voltage that gradually chokes off the current. Eventually, as t→∞, the capacitor becomes fully charged, and the current drops to zero. At this steady state, the voltage across the capacitor perfectly balances the motional EMF. The maximum charge stored is simply the capacitance multiplied by the EMF:
Qmax=C0E=(5.0×10−6)×(1.68×10−6)
This confirms that Option (C) is also correct. The beauty of this problem lies in how it seamlessly bridges the gap between electromagnetic induction and transient circuit analysis!
