The magnetic flux through a single turn coil with a resistance of 1 ohm changes: it decreases evenly in T1 time per second, and it decreases evenly in T2 time per second When the magnetic flux through a single turn coil with a resistance of 1 ohm changes: in T1 time, it decreases uniformly every second, and in T2 time, it increases uniformly every second, then the average value of the induced electromotive force generated in the coil must be zero in (T1 + T2) time. Why is it right? Let's not say that T1 = T2

The magnetic flux through a single turn coil with a resistance of 1 ohm changes: it decreases evenly in T1 time per second, and it decreases evenly in T2 time per second When the magnetic flux through a single turn coil with a resistance of 1 ohm changes: in T1 time, it decreases uniformly every second, and in T2 time, it increases uniformly every second, then the average value of the induced electromotive force generated in the coil must be zero in (T1 + T2) time. Why is it right? Let's not say that T1 = T2

Yes. Strictly speaking, it should be like this: "in T1 time, it decreases uniformly every second, and in T2 time, it increases uniformly at the same speed every second. If T1 = T2, the average value of the induced electromotive force generated in the coil in (T1 + T2) time must be zero."

There is a coil with 100 turns. The magnetic flux passing through the coil increases from 0.01wb to 0.09wb in 0.4s. The coil resistance is 10 ohm. The induced electromotive force and magnetic flux in the coil increase from 0.01wb to 0.09wb

The EMF generated by the coil is e = n * Δ Φ / Δ t = n * (Φ 2 - Φ 1) / Δ t = 100 * (0.09 - 0.01) / 0.4 = 20 v
The induced current in the coil is I = E / r = 20 / 10 = 2 A

There is a coil with n = 500 turns. The magnetic flux passing through it increases from 0.01wb to 0.09wb in 0.4s. The resistance of the coil is known to be 1000 ohm
Find: (1) the magnitude of the induced electromotive force in the coil (2) the current intensity in the coil

(1) The induced electromotive force E = n Δ Φ / Δ T, e = 500 × 0.08 / 0.4 = 100V
(2) I=E/R =100/1000=0.1A

If the flux passing through a single turn is always increased by 2 WB per second evenly, then ()
A. The induced electromotive force in the coil increases by 2VB per second. The induced electromotive force in the coil decreases by 2VC per second. The induced electromotive force in the coil is always 2VD. There is no induced electromotive force in the coil

According to Faraday's law of electromagnetic induction, e = n △ T, e = 2V remains unchanged. Therefore, C is correct and a, B and D are wrong

There is a coil with 100 turns, and the magnetic flux passing through it increases from 0.01wb to 0.09wb in 0.4s. The resistance of the coil is not included, but the resistance of the coil connection
The resistance value of the coil connection is 10 ohm, 1: how much is the induced electromotive force in the coil? 2: What is the thermal power of the resistor?

(1)E=n△φ/△t=100*(0.09Wb-0.01Wb)/0.4s=20V
(2)I=E/R=20V/10Ω=2A
P=I＾2*R=40W

In 0.4 seconds, the magnetic flux through a 10 turn closed coil increases from 0.02 WB to 0.08 WB, the coil resistance is 10 ohm, and the coil generates current intensity

Induced potential E = - △ ψ / △ t = - w △ Φ / △ t = - 10 × (0.08-0.02) / 0.4 = - 1.5V
Induced current I = E / r = - 1.5/10 = - 0.15A
The negative sign above indicates the direction to prevent the change of the magnetic field, the + sign indicates the direction of the right helix of the potential current and magnetic field, and the - sign indicates the opposite

There is a coil with 100 turns, and the magnetic flux passing through it increases from 0.01wb to 0.09wb in 0.4s, regardless of the resistance of the coil. The resistance of the coil connection is 10 Ω, which is calculated as follows: 1. The induced electromotive force in the coil; 2. The thermal power of the resistance

First, calculate the induced electromotive force, e = 100 (0.09-0.01) / 0.4 = 20V
Regardless of the coil resistance, there is no internal resistance, the external resistance voltage is 20V, the current is 20 / 10 = 2A, and the power is 20 * 2 = 40W

There is a coil with 1000 turns, and the magnetic flux passing through it increases from 0.02wb to 0.04wb in 0.4s. Find the induced electromotive force in the coil. If the resistance of the coil is 10 ohm, when it forms a closed loop with an electric heater with a resistance of 990 ohm, what is the current passing through the electric heater

According to Faraday's law of electromagnetic induction: e = n △ φ / △ t = 1000 × (0.04-0.02) / 0.4 = 50V;
According to Ohm's law of closed circuit: I = E / r = 50 / (10 + 990) = 0.05A

There is a coil with 1000 turns, the magnetic flux passing through it increases from 0.02wb to 0.09wb in 0.4s, and the induced electromotive force in the coil is calculated. If the resistance of the coil is 10 Ω, when it is connected in series with a resistance of 990 Ω to form a closed circuit, what is the current passing through the resistance?

Induced electromotive force E = 1000 x (0.09-0.02) / 0.4 = 175 (V);
Current I = E / (10 + 990) = 175 / 1000 = 0.17 (a)

There is a coil with 1000 turns. The magnetic flux passing through it increases from 0.02wb to 0.08wb in 0.3s. Find the induced electromotive force in the coil. If it is connected to a closed circuit with a total resistance of 1000 ohm, what is the current in the circuit?

Induced electromotive force in coil
E=n△ф/△t=1000*(0.08-0.02)/0.3=200V
Current in a circuit
I=E/R=200/1000=0.2A