Two spherical stars A and B have densities ρ_A and ρ_B, respectively. They have the same radius, and their masses M_A and M_B satisfy M_B = 2M_A. Star A loses mass such that its radius halves while ρ_A remains constant. The lost mass forms a spherical shell around B with density ρ_A. If v_A and v_B are the escape velocities after this process, and (v_B/v_A) = √(10n/15^(1/3)), find n.

 

Understanding the Problem

We have an LC circuit with:

• Inductance, $L=0.1$ H

• Capacitance, $C=10^{-3}$ F

• Area of the circuit, $A=1$ m²

This circuit is placed in a magnetic field given by:

$$B=B_0+\beta t$$

where $\beta =0.04$ T/s.

We need to find the maximum current (in mA) in this circuit.

Key Concepts Involved

1. LC Circuit: An LC circuit consists of an inductor (L) and a capacitor (C). It can oscillate with a natural frequency $\omega =\frac{1}{\sqrt{LC}}$.

2. Magnetic Flux: The magnetic flux through the circuit is $\mathrm{\Phi }=B\cdot A$. Since the magnetic field is changing with time, the flux is also changing.

3. Faraday's Law of Induction: A changing magnetic flux induces an electromotive force (emf) in the circuit:

   $$\mathcal{E}=-\frac{d\mathrm{\Phi }}{dt}$$

4. Induced Current: The induced emf will drive a current in the LC circuit, causing oscillations.

5. Maximum Current: In an LC circuit, the maximum current occurs when all the energy is in the inductor's magnetic field.

Step-by-Step Solution

Step 1: Calculate the Rate of Change of Magnetic Flux

Given:

$$B=B_0+\beta t$$$$\mathrm{\Phi }=B\cdot A=(B_0+\beta t)\cdot A$$$$\frac{d\mathrm{\Phi }}{dt}=\beta \cdot A$$

The negative sign indicates direction, but for magnitude, we can consider:

$$\mathcal{E}=0.04\text{ V}$$

Step 3: Effect of Induced emf on the LC Circuit

The induced emf acts as a driving source for the LC circuit. The LC circuit will oscillate with this driving emf. The maximum current in the circuit can be found by considering the energy imparted by the emf.

The energy stored in the capacitor due to the induced emf is:

$$U=\frac{1}{2}C{\mathcal{E}}^2$$

At maximum current, all this energy is transferred to the inductor:

$$\frac{1}{2}L{I}_{\text{max}}^2=\frac{1}{2}C{\mathcal{E}}^2$$$$I_{\text{max}}=\mathcal{E}\sqrt{\frac{C}{L}}$$

Step 4: Calculate Maximum Current

Plugging in the values:

$$\mathcal{E}=0.04\text{ V}$$$$C=10^{-3}\text{ F}$$$$L=0.1\text{ H}$$$$I_{\text{max}}=0.04\times \sqrt{\frac{10^{-3}}{0.1}}$$$$\sqrt{\frac{10^{-3}}{0.1}}=\sqrt{10^{-2}}=0.1$$$$I_{\text{max}}=0.04\times 0.1=0.004\text{ A}$$$$I_{\text{max}}=4\text{ mA}$$