To understand the magnetic induction at the center of two concentric circular conductors connected to different batteries, we need to break down the problem into manageable parts. We will analyze the magnetic field produced by each conductor and then compare the results.

Understanding the Setup

We have two circular conductors:

• The inner conductor with radius \( r_1 = 5 \, \text{cm} \) and resistance \( R_1 = 4 \, \Omega \).
• The outer conductor with radius \( r_2 = 10 \, \text{cm} \).

These conductors are connected to batteries with electromotive forces (EMFs) \( E_1 = 20 \, \text{V} \) for the inner conductor and \( E_2 = 16 \, \text{V} \) for the outer conductor.

Calculating Current in Each Conductor

Using Ohm's law, we can find the current flowing through each conductor:

• For the inner conductor:

  The current \( I_1 \) can be calculated as:

  I_1 = \frac{E_1}{R_1} = \frac{20 \, \text{V}}{4 \, \Omega} = 5 \, \text{A}

• For the outer conductor, we need to know its resistance. Assuming it has the same thickness and material, we can find its resistance using the formula for resistance in terms of resistivity, length, and area. However, for simplicity, let’s denote its resistance as \( R_2 \) and find the current \( I_2 \) as follows:

I_2 = \frac{E_2}{R_2}

Magnetic Field at the Center

The magnetic field \( B \) at the center of a circular loop carrying current can be calculated using the formula:

B = \frac{\mu_0 I}{2r}

Where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)), \( I \) is the current, and \( r \) is the radius of the loop.

Magnetic Field from Inner Conductor

For the inner conductor:

B_1 = \frac{\mu_0 I_1}{2r_1} = \frac{4\pi \times 10^{-7} \times 5}{2 \times 0.05} = \frac{4\pi \times 10^{-7} \times 5}{0.1} = 2\pi \times 10^{-6} \, \text{T}

Magnetic Field from Outer Conductor

For the outer conductor, we need to calculate the current \( I_2 \) first. Assuming \( R_2 \) is proportional to the square of the radius (since resistance increases with length), we can express it as:

R_2 = k \cdot r_2^2

Thus, the current \( I_2 \) can be expressed as:

I_2 = \frac{E_2}{k \cdot r_2^2}

Substituting this into the magnetic field formula:

B_2 = \frac{\mu_0 I_2}{2r_2} = \frac{\mu_0 \cdot \frac{E_2}{k \cdot r_2^2}}{2r_2} = \frac{\mu_0 E_2}{2k \cdot r_2^3}

Comparing Magnetic Inductions

Now, we can find the ratio of the magnetic inductions at the center from both conductors:

\frac{B_1}{B_2} = \frac{2\pi \times 10^{-6}}{\frac{\mu_0 E_2}{2k \cdot r_2^3}}

This ratio will depend on the specific values of \( k \) and \( r_2 \). However, we can see that the magnetic field strength is influenced by the current and the radius of the conductors.

Conclusion

In summary, the magnetic induction at the center of the two concentric circular conductors can be calculated using the currents derived from their respective EMFs and resistances. The magnetic fields produced by each conductor can then be compared to find their ratio. This analysis highlights the relationship between current, radius, and magnetic induction in circular loops.