JEE Challenger
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PhyHardAdv 2021 • Paper 1

Motional EMF on Semicircular Rod Near Current Wire Connected to RC Circuit

A long straight wire carries a current I=2 ampereI = 2\text{ ampere}. A semi-circular conducting rod rests on two parallel conducting rails of negligible resistance, lying in the same horizontal plane as the wire. The two ends of the rod are situated at distances of 1 cm1\text{ cm} and 4 cm4\text{ cm} from the long wire. Starting at time t=0t = 0, the rod moves along the rails at a velocity of v=3.0 m/sv = 3.0\text{ m/s} (refer to figure). A resistor R=1.4 ΩR = 1.4\ \Omega and an uncharged capacitor Co=5.0 μFC_o = 5.0\ \mu\text{F} are connected in series across the rails. Which of the following statement(s) is(are) correct? [Given: μ0=4π×10−7 SI units\mu_0 = 4\pi \times 10^{-7}\text{ SI units}, ln⁡2=0.7\ln 2 = 0.7]

Question Diagram 1

Options

A

Maximum current through RR is 1.2×10−6 ampere1.2 \times 10^{-6}\text{ ampere}

Correct
B

Maximum current through RR is 3.8×10−6 ampere3.8 \times 10^{-6}\text{ ampere}

C

Maximum charge on capacitor CoC_o is 8.4×10−12 coulomb8.4 \times 10^{-12}\text{ coulomb}

Correct
D

Maximum charge on capacitor CoC_o is 2.4×10−12 coulomb2.4 \times 10^{-12}\text{ coulomb}

Topics & Concepts

Electromagnetic InductionFaraday's LawLenz's LawCapacitors

Step-by-Step Solution

To determine the correct options, we analyze the motional electromotive force (EMF) induced across the moving semi-circular rod and its effect on the series RCRC circuit.

1. Calculation of Induced Motional EMF (E\mathcal{E})

The magnetic field B⃗\vec{B} at a distance xx from a long straight wire carrying current II is directed perpendicular to the plane containing the wire and the rails (into or out of the plane) and has magnitude: B(x)=μ0I2πxB(x) = \frac{\mu_0 I}{2\pi x}

The semi-circular rod moves with a constant speed vv along the rails, parallel to the straight wire. The induced motional EMF across the ends of any conducting path moving through a magnetic field is given by: E=∫(v⃗×B⃗)⋅dℓ⃗\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{\ell}

Since v⃗\vec{v} is directed along the rails (parallel to the wire) and B⃗\vec{B} is perpendicular to the plane, v⃗×B⃗\vec{v} \times \vec{B} points radially away from or towards the wire (perpendicular to the wire). Therefore, only the radial distance component dxdx contributes to the total EMF, making the shape of the rod irrelevant except for its end positions x1=1 cm=0.01 mx_1 = 1\text{ cm} = 0.01\text{ m} and x2=4 cm=0.04 mx_2 = 4\text{ cm} = 0.04\text{ m}: E=∫x1x2vB(x) dx=∫x1x2v(μ0I2πx)dx=μ0Iv2πln⁡(x2x1)\mathcal{E} = \int_{x_1}^{x_2} v B(x) \, dx = \int_{x_1}^{x_2} v \left(\frac{\mu_0 I}{2\pi x}\right) dx = \frac{\mu_0 I v}{2\pi} \ln\left(\frac{x_2}{x_1}\right)

Given parameters:

  • μ0=4π×10−7 N/A2\mu_0 = 4\pi \times 10^{-7}\text{ N/A}^2
  • I=2 AI = 2\text{ A}
  • v=3.0 m/sv = 3.0\text{ m/s}
  • x2x1=4 cm1 cm=4\frac{x_2}{x_1} = \frac{4\text{ cm}}{1\text{ cm}} = 4
  • ln⁡2=0.7  ⟹  ln⁡4=2ln⁡2=1.4\ln 2 = 0.7 \implies \ln 4 = 2\ln 2 = 1.4

Substituting these values into the expression for E\mathcal{E}: E=(4π×10−7)(2)(3.0)2πln⁡(4)=(4×10−7)×3.0×1.4=1.68×10−6 V\mathcal{E} = \frac{(4\pi \times 10^{-7})(2)(3.0)}{2\pi} \ln(4) = (4 \times 10^{-7}) \times 3.0 \times 1.4 = 1.68 \times 10^{-6}\text{ V}

Since vv is constant and the distance of the rails from the wire remains constant, E=1.68×10−6 V\mathcal{E} = 1.68 \times 10^{-6}\text{ V} behaves as a constant DC voltage source in the circuit.

2. Maximum Current through the Resistor RR

In a series RCRC circuit connected to a constant DC EMF source E\mathcal{E}, the current at any time tt is given by: i(t)=ERe−t/RCoi(t) = \frac{\mathcal{E}}{R} e^{-t/RC_o}

The maximum current occurs at t=0+t = 0^+ immediately after the rod starts moving when the capacitor is uncharged (q=0q=0): imax=ERi_{\text{max}} = \frac{\mathcal{E}}{R}

Given R=1.4 ΩR = 1.4\ \Omega: imax=1.68×10−6 V1.4 Ω=1.2×10−6 Ai_{\text{max}} = \frac{1.68 \times 10^{-6}\text{ V}}{1.4\ \Omega} = 1.2 \times 10^{-6}\text{ A}

This confirms that Statement (A) is correct.

3. Maximum Charge on Capacitor CoC_o

As t→∞t \to \infty, the current drops to zero and the capacitor becomes fully charged. The potential difference across the capacitor equals the induced EMF E\mathcal{E}: qmax=CoEq_{\text{max}} = C_o \mathcal{E}

Given Co=5.0 μF=5.0×10−6 FC_o = 5.0\ \mu\text{F} = 5.0 \times 10^{-6}\text{ F}: qmax=(5.0×10−6 F)×(1.68×10−6 V)=8.4×10−12 Cq_{\text{max}} = (5.0 \times 10^{-6}\text{ F}) \times (1.68 \times 10^{-6}\text{ V}) = 8.4 \times 10^{-12}\text{ C}

This confirms that Statement (C) is correct.

Conclusion

  • (A) Maximum current through RR is 1.2×10−6 ampere1.2 \times 10^{-6}\text{ ampere} (Correct)
  • (B) Maximum current through RR is 3.8×10−6 ampere3.8 \times 10^{-6}\text{ ampere} (Incorrect)
  • (C) Maximum charge on capacitor CoC_o is 8.4×10−12 coulomb8.4 \times 10^{-12}\text{ coulomb} (Correct)
  • (D) Maximum charge on capacitor CoC_o is 2.4×10−12 coulomb2.4 \times 10^{-12}\text{ coulomb} (Incorrect)

Correct Options: A, C

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