In this problem, we are tasked with determining the magnetic flux and the average current in the circuit for a time interval $5 \, \text{seconds} < t < 10 \, \text{seconds}$. The induced electromotive force (emf) is given as $e(t) = 2t + 6$, and we know the resistance values of the circuit components. Here is how we proceed step by step:

Given Information:

• Number of turns in the coil, $N = 100$
• Coil resistance, $R_{\text{coil}} = 100 \, \Omega$
• Galvanometer resistance, $R_{\text{galv}} = 500 \, \Omega$
• Total resistance in the circuit, $R = R_{\text{coil}} + R_{\text{galv}} = 100 + 500 = 600 \, \Omega$
• Induced emf as a function of time, $e(t) = 2t + 6$
• Time interval: $t = 5 \, \text{seconds}$ to $t = 10 \, \text{seconds}$

Step 1: Calculate the Magnetic Flux

Faraday's Law of Electromagnetic Induction states that: $e(t) = -N \frac{d\Phi_B}{dt}$ Where:

• $N$ is the number of turns in the coil.
• $\Phi_B$ is the magnetic flux through the coil.

Rearranging the formula for magnetic flux: $\frac{d\Phi_B}{dt} = -\frac{e(t)}{N}$ Substituting $e(t) = 2t + 6$ and $N = 100$: $\frac{d\Phi_B}{dt} = -\frac{2t + 6}{100}$ Now, we can integrate this expression over time to find the magnetic flux $\Phi_B$. The integral of $\frac{d\Phi_B}{dt}$ will give us $\Phi_B(t)$.

The initial flux at $t = 5$ seconds and the final flux at $t = 10$ seconds will be calculated from this expression.

Step 2: Calculate the Average Current

The current in the circuit is related to the induced emf by Ohm’s Law: $I(t) = \frac{e(t)}{R}$ Where:

• $e(t) = 2t + 6$
• $R = 600 \, \Omega$

To find the average current over the time interval $5 \, \text{seconds} < t < 10 \, \text{seconds}$, we calculate: $I_{\text{avg}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} I(t) \, dt$ Where:

• $t_1 = 5 \, \text{seconds}$
• $t_2 = 10 \, \text{seconds}$
• $I(t) = \frac{2t + 6}{600}$

We will now compute the results step by step.

Let's start with the magnetic flux calculation and then move on to the current.### Results:

1. Magnetic Flux:

   • Magnetic flux at $t = 5$ seconds: $\Phi_5 = -\frac{11}{20} \, \text{Wb} = -0.55 \, \text{Wb}$
   • Magnetic flux at $t = 10$ seconds: $\Phi_10 = -\frac{8}{5} \, \text{Wb} = -1.6 \, \text{Wb}$
   • Change in magnetic flux over the time interval: $\Delta \Phi = -\frac{21}{20} \, \text{Wb} = -1.05 \, \text{Wb}$

2. Average Current:

   • The average current in the circuit over the interval $5 \, \text{seconds} < t < 10 \, \text{seconds}$ is $I_{\text{avg}} = 0.035 \, \text{A}$ or 35 mA.

Summary:

• The magnetic flux changes by $-1.05 \, \text{Wb}$ during the time interval.
• The average current induced in the circuit is 35 mA.

Would you like further details or have any questions?

Relative Questions:

1. How does the number of turns in the coil affect the induced emf?
2. What is the significance of the negative sign in the change of magnetic flux?
3. How would increasing the resistance in the circuit impact the average current?
4. Can we use this method to find the instantaneous current at a specific time?
5. How does Faraday's Law of Induction relate to Lenz's Law?

Tip:

Always remember that increasing resistance in the circuit will reduce the induced current for the same emf, as per Ohm’s Law.