The magnetic field in the region shown below is increasing by 5 T/s. The circuit shown (with R=25 , C=55 µF, dimensions 1 m by 2 m) is inserted into the region as shown. The capacitor is initially uncharged.

1. What is the current in the circuit at the moment the circuit is put in the B field?
2. How long will it take the capacitor to charge to 25% of its final charge?
3. How much charge will the capacitor hold after a long time (assuming this magnetic field rate increase is maintained)?

Please help i don't understand which formulas i need to use

To solve these questions, you will need to use the concepts of Faraday's law of electromagnetic induction, Ohm's law, and the equation for the charging of a capacitor. The formulas you will need are:

1. Faraday's law of electromagnetic induction: EMF = -d(Φ)/dt
- EMF is the electromotive force induced in the circuit
- Φ is the magnetic flux through the circuit
- t is time

2. Ohm's law: V = IR
- V is the voltage across the resistor
- I is the current flowing through the circuit
- R is the resistance of the resistor

3. Equation for charging a capacitor: Q = CV
- Q is the charge stored in the capacitor
- C is the capacitance of the capacitor
- V is the voltage across the capacitor

Now, let's solve each question step by step:

1. What is the current in the circuit at the moment the circuit is put in the B field?

At the moment the circuit is inserted into the magnetic field, there will be an induced EMF according to Faraday's law. This EMF will cause a current to flow through the circuit. The induced EMF can be calculated using the formula:

EMF = -d(Φ)/dt

Since the magnetic field is increasing with a rate of 5 T/s, we can calculate the change in the magnetic flux by multiplying the rate of change of the magnetic field by the area of the circuit (1 m by 2 m):

Φ = B * A

Φ = (5 T/s) * (1 m * 2 m) = 10 T·m²/s

Now, we can substitute the values into the equation for the induced EMF:

EMF = -d(Φ)/dt = -10 T·m²/s / dt

Since the capacitor is initially uncharged, the current will be maximum at the initial moment. Therefore, the current in the circuit at that moment is:

I = EMF / R = (-10 T·m²/s) / 25 Ω

2. How long will it take the capacitor to charge to 25% of its final charge?

The time taken for the capacitor to charge can be determined using the formula for charging a capacitor:

Q = CV

At 25% of its final charge, the charge held by the capacitor is 0.25 times the maximum charge it can hold. We can express this as:

Q = 0.25 * (C * Vmax)

Rearranging the equation, we get:

Vmax = Q / (0.25 * C)

The time taken for the capacitor to charge can be calculated using the formula:

t = RC * ln(1 - Q / (C * Vmax))

where ln is the natural logarithm.

3. How much charge will the capacitor hold after a long time (assuming this magnetic field rate increase is maintained)?

Since the circuit is inserted into a magnetic field with a continuously increasing rate, a constant current will flow through the circuit. As time passes, the capacitor will charge until it reaches its maximum charge.

The maximum charge that the capacitor can hold is given by the equation for charging a capacitor:

Qmax = C * Vmax

Using the given values, you can calculate the maximum charge held by the capacitor.

To answer these questions, we can use the principles of electromagnetic induction and RC circuits.

1. To find the current in the circuit at the moment it is put in the magnetic field, we can apply Faraday's law of electromagnetic induction. This law states that the induced electromotive force (EMF) in a circuit is proportional to the rate of change of magnetic flux through the circuit. Mathematically, it can be expressed as:

EMF = -d(Φ)/dt

Where EMF is the induced electromotive force, Φ is the magnetic flux, and dt represents the change in time. In this case, the magnetic field is increasing by 5 T/s, which means the change in magnetic flux is 5 T multiplied by the area of the circuit.

Φ = B * A

Where B is the magnetic field and A is the area of the circuit. Given the dimensions of 1 m by 2 m, the area is 1 m x 2 m = 2 m^2.

Now we can calculate the induced EMF:

EMF = -d(Φ)/dt = -d(B * A)/dt = -A * dB/dt

Plugging in the values, we have:

EMF = -2 m^2 * (5 T/s) = -10 V

Since the capacitor is initially uncharged, it acts as a short circuit. Therefore, the current in the circuit at the moment it is put in the magnetic field is given by Ohm's law:

I = V/R

Where V is the voltage (EMF) and R is the resistance. From the above calculation, we have:

I = -10 V / 25 Ω = -0.4 A (Note: The negative sign indicates the direction of the current.)

So, the current in the circuit at that moment is 0.4 A in the opposite direction to the increase in the magnetic field.

2. To calculate the time it takes for the capacitor to charge to 25% of its final charge, we need to use the time constant (τ) of an RC circuit. The time constant is given by the formula:

τ = R * C

Where R is the resistance and C is the capacitance. In this case, R = 25 Ω and C = 55 µF (or 55 x 10^-6 F).

Using the given values, we have:

τ = (25 Ω) * (55 x 10^-6 F) = 1.375 ms

The time constant represents the time it takes for the charge of the capacitor to reach approximately 63.2% of its final charge. To find the time it takes to reach 25% of the final charge, we can use the formula for the charging (or discharging) of a capacitor in an RC circuit:

Q(t) = Q_max * (1 - e^(-t/τ))

Where Q(t) is the charge of the capacitor at time t, Q_max is the maximum charge, t is the time, and e is the base of the natural logarithm.

We want to find the time (t) when Q(t) = 0.25 * Q_max, so we can rearrange the formula:

0.25 * Q_max = Q_max * (1 - e^(-t/τ))

Simplifying the equation:

0.25 = 1 - e^(-t/τ)

Rearranging further:

e^(-t/τ) = 1 - 0.25

e^(-t/τ) = 0.75

To solve for t, take the natural logarithm (ln) of both sides:

-t/τ = ln(0.75)

t/τ = -ln(0.75)

t = -τ * ln(0.75)

Substituting the value of τ calculated earlier:

t = -1.375 ms * ln(0.75) ≈ -1.375 ms * (-0.2877) ≈ 0.396 ms

Therefore, it takes approximately 0.396 ms for the capacitor to charge to 25% of its final charge.

3. To determine the charge the capacitor holds after a long time, we assume that the magnetic field rate increase is maintained indefinitely. In this case, the capacitor will reach its final charge. In a fully charged capacitor, the charge (Q) is given by:

Q = Q_max = V_max * C

Where V_max is the maximum voltage across the capacitor and C is the capacitance.

In an RC circuit, V_max can be calculated using Ohm's law:

V_max = I_max * R

Where I_max is the maximum current in the circuit. Since the capacitor is fully charged, there is no current flowing through the capacitor and the resistor. Therefore:

I_max = 0 A

Substituting this into the equation for V_max:

V_max = 0 A * R = 0 V

Thus, the charge the capacitor holds after a long time is:

Q = Q_max = V_max * C = 0 V * 55 µF = 0 C

Therefore, the capacitor will hold zero charge after a long time when the magnetic field rate increase is maintained.