Question 1

What is the self-inductance of an ideal solenoid that is 300 cm long with a cross-sectional area of 1.00 × 10^-4 m² and has 1000 turns of wire? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

The self-inductance L of a solenoid is given by the formula L = (μ₀ * N² * A) / l, where N is the number of turns, A is the cross-sectional area, and l is the length. Substituting the given values, L = (4π × 10⁻⁷ T∙m/A * 1000² * 1.00 × 10⁻⁴ m²) / 3.00 m = 41.9 µH.

Question 2

An ideal step-up transformer doubles a primary voltage of 110 V.What is the ratio of the number of turns in its primary coil to the number of turns in the secondary coil?

Accepted Answer

A step-up transformer increases voltage, which means the secondary coil must have more turns than the primary coil. If it doubles the voltage from 110 V to 220 V, the ratio of primary to secondary turns is 1:2, indicating the secondary has twice as many turns as the primary.

Question 3

The inductance of a solenoid that is 16.0 cm long and has a cross-sectional area of 1.00 × Is How many turns of wire does this solenoid have? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

The answer of The inductance of a solenoid that is...

Question 4

The primary of an ideal transformer has 100 turns and its secondary has 200 turns.If the input voltage to the primary is 100 V,we can expect the output voltage of the secondary to be

Accepted Answer

The turns ratio of the transformer is 1:2. Therefore, the output voltage will be twice the input voltage. So, the output voltage of the secondary will be 200 V.

Question 5

In an ideal transformer,how many turns are necessary in a 110-V primary if the 24-V secondary has 100 turns?

Accepted Answer

The turns ratio in a transformer is directly proportional to the voltage ratio. Therefore, if the secondary has 100 turns and produces 24 V, for a primary voltage of 110 V, the number of turns in the primary would be calculated as follows: $ \frac{110V}{24V} = \frac{N_{primary}}{100 turns} $, solving for $ N_{primary} $ gives $ N_{primary} = \frac{110}{24} \times 100 = 458.33 $ turns. Since the number of turns must be an integer, we round to the nearest whole number, which is 458.

Question 6

A generator produces 60 A of current at 120 V.The voltage is usually stepped up to 4500 V by an ideal transformer and transmitted through a power line of total resistance 1.0 Ω.Find the number of turns in the secondary if the primary has 200 turns.

Accepted Answer

The number of turns in the secondary can be found using the transformer equation, which relates the primary and secondary voltages (Vp and Vs) to the number of turns in the primary (Np) and secondary (Ns) coils: Vs/Vp = Ns/Np. Given Vp = 120 V, Vs = 4500 V, and Np = 200 turns, solving for Ns gives Ns = (Vs/Vp) * Np = (4500/120) * 200 = 7500 turns.

Question 7

A coil with a self-inductance of 6.0 H is connected to a dc battery through a switch.As soon as the switch is closed,the rate of change of current is 2.0 A/s.What is the emf induced in this coil at this instant?

Accepted Answer

The emf induced in the coil at any instant is given by the formula emf = -L*(di/dt), where L is the self-inductance of the coil and di/dt is the rate of change of current.

Substituting the given values, we have:

emf = -6.0 H * (2.0 A/s) = -12 V

Since the emf is induced in the opposite direction to the current, the negative sign indicates that the emf opposes the increase in current. Therefore, the magnitude of the emf is 12 V.

Hence, the correct answer is C) 12V

Question 8

The primary of an ideal transformer has 100 turns and its secondary has 200 turns.Neglecting frictional losses,if the power input to the primary is 100 W,we can expect the power output of the secondary to be

Accepted Answer

In an ideal transformer, the power output of the secondary is equal to the power input of the primary (neglecting losses), so if the primary input is 100 W, the secondary output is also 100 W.

Question 9

The primary coil of an ideal transformer has 600 turns and its secondary coil has 150 turns.If the current in the primary coil is 2 A,what is the current in its secondary coil?

Accepted Answer

The current in the secondary coil can be found using the transformer equation $I_1/I_2 = N_2/N_1$, where $I_1$ and $I_2$ are the currents in the primary and secondary coils, respectively, and $N_1$ and $N_2$ are the number of turns in the primary and secondary coils, respectively. Substituting the given values, $2/I_2 = 150/600$, solving for $I_2$ gives $I_2 = 8$ A.

Question 10

An ideal transformer consists of a 500-turn primary coil and a 2000-turn secondary coil.If the current in the secondary is 3.0 A,what is the current in the primary?

Accepted Answer

The current in the primary coil can be found using the transformer equation $I_p = \frac{N_s}{N_p} \times I_s$, where $I_p$ is the primary current, $N_s$ and $N_p$ are the number of turns in the secondary and primary coils respectively, and $I_s$ is the secondary current. Substituting the given values, $I_p = \frac{2000}{500} \times 3.0 A = 4 \times 3.0 A = 12 A$.

Question 11

The primary coil of an ideal transformer has 100 turns and its secondary coil has 400 turns.If the ac current in the secondary coil is 2 A,what is the current in its primary coil?

Accepted Answer

The answer of The primary coil of an ideal transformer...

Question 12

A coil with a self-inductance of 6.0 H has a constant current of 2.0 A flowing through it for 2.0 s.What is the emf induced in this coil?

Accepted Answer

The emf induced in a coil is given by Faraday's law of electromagnetic induction, which states that the induced emf is proportional to the rate of change of magnetic flux. Since the current through the coil is constant, there is no change in magnetic flux, and therefore, no emf is induced.

Question 13

A current of 2.0 A in the 100-turn primary of an ideal transformer causes 14 A to flow in the secondary.How many turns are in the secondary?

Accepted Answer

The answer of A current of 2.0 A in the...

Question 14

The primary of an ideal transformer has 100 turns and its secondary has 200 turns.If the input current at the primary is 100 A,we can expect the output current at the secondary to be

Accepted Answer

The output current can be calculated using the transformer turns ratio, which is the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. For an ideal transformer, the input power equals the output power, and the turns ratio inversely affects the current. Given a turns ratio of 100:200 (or 1:2), if the primary current is 100 A, the secondary current will be half of that, which is 50 A.

Question 15

When 5.0 A at 110 V flows in the primary of an ideal transformer,how many amps at 24 V can flow in the secondary?

Accepted Answer

In an ideal transformer, the power in the primary coil (P1) equals the power in the secondary coil (P2). Power is calculated as P = IV, where I is current and V is voltage. For the primary coil, P1 = 110V * 5.0A = 550W. To find the current in the secondary coil with a voltage of 24V, we use P2 = P1 = 550W, so I2 = P2 / V2 = 550W / 24V = 22.9167A, which rounds to 23A, making option D correct.

Question 16

An ideal solenoid with 3000 turns is 70.0 cm long.If its self-inductance is 25.0 mH,what is its radius? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

The self-inductance $L$ of a solenoid is given by $L = \mu_0 \frac{N^2 A}{l}$, where $N$ is the number of turns, $A$ is the cross-sectional area, $l$ is the length of the solenoid, and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, T \cdot m/A$). Rearranging for $A$ and substituting $L$, $N$, $l$, and $\mu_0$ into the equation, and then solving for the radius $r$ (since $A = \pi r^2$), we find that the radius that matches the given self-inductance is 0.0222 m.

Question 17

The secondary coil of an ideal neon sign transformer provides 7500 V at 10.0 mA.The primary coil operates on 120 V.What current does the primary draw?

Accepted Answer

The transformer's power on both sides (primary and secondary) is equal in an ideal transformer. So, $P_{primary} = P_{secondary}$. Given $V_{secondary} = 7500V$ and $I_{secondary} = 10.0mA = 0.01A$, $P_{secondary} = V_{secondary} \times I_{secondary} = 7500V \times 0.01A = 75W$. With $V_{primary} = 120V$, the primary current $I_{primary} = \frac{P_{primary}}{V_{primary}} = \frac{75W}{120V} = 0.625A$.

Question 18

The current flowing through a circuit is changing at a rate of 6.0 A/s.If the circuit contains a 190-H inductor,what is the emf across the inductor?

Accepted Answer

emf = L di/dt = (190 H)(6.0 A/s) = 1140 V

Quiz 23: Magnetic Flux and Faradays Law of Induction

What is the self-inductance of an ideal solenoid that is 300 cm long with a cross-sectional area of 1.00 × 10-4 m2 and has 1000 turns of wire? (μ0 = 4π × 10-7 T ∙ m/A)