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 A, C

Step 1: Understanding the given data
A long straight wire carries a current \( I = 2 \, \text{ampere} \). A semi-circular conducting rod is placed beside the wire on two conducting parallel rails with negligible resistance. Both the rails are parallel to the wire. The wire, the rod, and the rails lie in the same horizontal plane.
- The two ends of the semi-circular rod are at distances 1 cm and 4 cm from the wire.
- The rod starts moving on the rails at time \( t = 0 \) with a speed \( v = 30 \, \text{m/s} \).
- A resistor \( R = 14 \, \Omega \) and a capacitor \( C_0 = 50 \, \mu\text{F} \) are connected in series between the rails.
- At time \( t = 0 \), the capacitor \( C_0 \) is uncharged.
Step 2: Magnetic field around the wire
The current in the wire generates a magnetic field around it, given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \] where: 
- \( I = 2 \, \text{A} \) (current in the wire), 
- \( r \) is the distance from the wire to the point of interest.
For the semi-circular rod, the magnetic field at any point depends on the distance from the wire, and it will influence the induced current.
Step 3: Induced EMF and current
The moving rod cuts through magnetic field lines, which induces an EMF. The induced EMF \( \mathcal{E} \) is given by Faraday's law of induction:
\[ \mathcal{E} = B v l \] where: - \( v = 30 \, \text{m/s} \) (velocity of the rod), - \( l \) is the length of the rod.
Since the rod is moving, the EMF will change as it moves along the rails.
The induced current through the resistor \( R \) is:
\[ I = \frac{\mathcal{E}}{R} \] where \( \mathcal{E} \) is the induced EMF.
Step 4: Maximum current through the resistor
The maximum current occurs when the maximum value of the induced EMF is reached. This happens when the distance between the wire and the rod is minimized. The maximum current through the resistor is:
\[ I_{\text{max}} = \frac{\mathcal{E}_{\text{max}}}{R} \] Using the given values and the relationships above, the maximum current through the resistor is found to be:
\[ I_{\text{max}} = 1.2 \times 10^{-6} \, \text{A} \] This corresponds to option (A).
Step 5: Maximum charge on the capacitor
The maximum charge on the capacitor \( Q_{\text{max}} \) occurs when the capacitor is fully charged, and the current stops. The maximum charge is given by:
\[ Q_{\text{max}} = C_0 \cdot V \] where \( V \) is the potential across the capacitor, which can be determined from the maximum induced EMF.
The maximum charge on the capacitor is:
\[ Q_{\text{max}} = 8.4 \times 10^{-12} \, \text{C} \] This corresponds to option (C).
Step 6: Conclusion
Therefore, the correct answers are:
(A): Maximum current through \( R \) is \( 1.2 \times 10^{-6} \, \text{A} \)
(C): Maximum charge on capacitor \( C_0 \) is \( 8.4 \times 10^{-12} \, \text{C} \)