Question

A long straight wire carries a current, $I =2$ ampere A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance Both the rails are parallel to the wire The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure Two ends of the semi-circular rod are at distances $1 \,cm$ and $4 \,cm$ from the wire At time $t =0$, the rod starts moving on the rails with a speed $v =30\, m / s$ (see the figure)
 A resistor $R =14 \,\Omega$ and a capacitor $C _{0}=50\, \mu F$ are connected in series between the rails At time $t =0, C _{0}$ is uncharged. Which of the following statement(s) is(are) correct? \([\mu_0 = 4 \pi \times 10^{-7}\) SI units. Take \(ln_2=0.7]\)

• A. Maximum current through $R$ is $1.2 \times 10^{-6}$ ampere
• B. Maximum current through $R$ is $3.8 \times 10^{-6}$ ampere
• C. Maximum charge on capacitor $C_{0}$ is $8.4 \times 10^{-12}$ coulomb
• D. Maximum charge on capacitor $C_{0}$ is $2.4 \times 10^{-12}$ coulomb

Answer 1

The Correct Option is C

Given:

• Current in wire, \( I = 2 \, \text{A} \)
• Semi-circular rod moves with velocity \( v = 30 \, \text{m/s} \)
• Distances of rod ends from wire: \( r_1 = 1 \, \text{cm} = 0.01 \, \text{m}, \quad r_2 = 4 \, \text{cm} = 0.04 \, \text{m} \)
• Capacitor \( C_0 = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} \), Resistor \( R = 14 \, \Omega \)
• \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \), \( \ln 2 = 0.7 \)

Step 1: Magnetic field at distance r from a long wire

\[ B(r) = \frac{\mu_0 I}{2\pi r} \]

Step 2: emf induced across the rod due to motion in magnetic field

Since the rod moves perpendicular to the field, the emf is: \[ \mathcal{E} = v \int_{r_1}^{r_2} B(r) \, dr = v \int_{r_1}^{r_2} \frac{\mu_0 I}{2\pi r} \, dr = v \cdot \frac{\mu_0 I}{2\pi} \cdot \ln\left( \frac{r_2}{r_1} \right) \] Substitute values: \[ \mathcal{E} = 30 \cdot \frac{4\pi \times 10^{-7} \cdot 2}{2\pi} \cdot \ln\left( \frac{0.04}{0.01} \right) = 30 \cdot 4 \times 10^{-7} \cdot \ln(4) \] \[ \ln 4 = \ln(2^2) = 2 \ln 2 = 2 \cdot 0.7 = 1.4 \] \[ \mathcal{E} = 30 \cdot 4 \times 10^{-7} \cdot 1.4 = 1.68 \times 10^{-5} \, \text{V} \]

Step 3: Maximum charge on capacitor

Maximum charge: \[ q_{\text{max}} = C_0 \cdot \mathcal{E} = 50 \times 10^{-6} \cdot 1.68 \times 10^{-5} = 8.4 \times 10^{-12} \, \text{C} \]

Correct Answer: Option (C): Maximum charge on capacitor is \( \boxed{8.4 \times 10^{-12}} \, \text{C} \)