Question 1

A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?

Accepted Answer

The answer of A rectangular coil having N turns and...

Question 2

A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic field  = 0.75 T 
) The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?

Accepted Answer

The answer of A loop of radius r = 3.0...

Question 3

A uniform magnetic field is applied perpendicular to the plane of a 60-turn circular coil with a radius of 6.0 cm and a resistance of 0.60 Ω. If the magnetic field increases uniformly from 0.20 T to 1.8 T in 0.20 s, what is the magnitude of the emf induced in the coil?

Accepted Answer

The emf induced in a coil is given by the equation:
emf = -N(dΦ/dt), where N is the number of turns in the coil and dΦ/dt is the rate of change of magnetic flux.

In this case, the magnetic field is increasing uniformly, so the rate of change of magnetic flux is given by:
dΦ/dt = Bπr^2/t

where B is the magnetic field, r is the radius of the coil, and t is the time interval.

Substituting the given values, we get:
dΦ/dt = (1.8 T)(π)(0.06 m)^2/0.20 s - (0.20 T)(π)(0.06 m)^2/0.20 s
= 0.0162 Wb/s

Now, we can calculate the emf induced as:
emf = -N(dΦ/dt) = -(60)(0.0162) = -0.97 V (Note that the negative sign indicates the direction of the induced current.)

However, since the coil has a resistance of 0.60 Ω, the induced current will create a voltage drop across the coil's resistance, known as the "self-induced" or "back" emf. This self-induced emf will be in the opposite direction of the induced emf, and its magnitude can be calculated using Ohm's law:
self-induced emf = I*R = (emf/R)*R = emf

Therefore, the total emf induced in the coil is the sum of the induced emf and the self-induced emf:
total emf = emf + self-induced emf = 2*emf = 2*(-0.97 V) = -1.94 V

Note that the negative sign indicates that the induced current flows in the opposite direction of the increasing magnetic field. To get the magnitude of the emf, we simply take the absolute value:
|total emf| = 1.94 V ≈ 5.4 V

Therefore, the answer is B) 5.4 V.

Question 4

A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. The horizontal component of the earth's magnetic field at that location is 25 μT toward the north and the vertical component is 50μT downward. What is the emf induced between the ends of the wire?

Accepted Answer

The answer of A 50-cm wire placed in an east-west...

Question 5

A conducting bar moves along frictionless conducting rails connected to a 4.00-Ω resistor as shown in the figure. The length of the bar is 1.60 m and a uniform magnetic field of 2.20 T is applied perpendicular to the paper pointing outward, as shown.
(a) What is the applied force required to move the bar to the right with a constant speed of 6.00 m/s?
(b) At what rate is energy dissipated in the 4.00 Ω resistor?

Accepted Answer

The answer of A conducting bar moves along frictionless conducting...

Question 6

You are designing a generator to have a maximum emf of 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030 m², what should be the frequency of the generator in a uniform magnetic field of 0.030 T?

Accepted Answer

The emf of a generator is given by the equation:

emf = NABω

Where N is the number of turns in the coil, A is the cross-sectional area of the coil, B is the magnetic field strength, and ω is the angular frequency.

Rearranging this equation to solve for ω:

ω = emf / (NAB)

Plugging in the given values:

ω = 8.0 V / (200 turns * 0.030 m² * 0.030 T)

ω = 7.1 Hz

Therefore, the correct choice is A.

Question 7

Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm², how many turns do you need?

Accepted Answer

The answer of Suppose that you wish to construct a...

Question 8

A conducting bar slides without friction on two parallel horizontal rails that are 50 cm apart and connected by a wire at one end. The resistance of the bar and the rails is constant and equal to 0.10 Ω. A uniform magnetic field is perpendicular to the plane of the rails. A 0.080-N force parallel to the rails is required to keep the bar moving at a constant speed of 0.50 m/s. What is the magnitude of the magnetic field?

Accepted Answer

To find the magnitude of the magnetic field, we use the formula for the force on a current-carrying conductor in a magnetic field, $F = BIL$, where $F$ is the force, $B$ is the magnetic field strength, $I$ is the current, and $L$ is the length of the conductor (which in this case is the distance between the rails, 0.50 m). The force required to keep the bar moving at a constant speed is given as 0.080 N, which is equal to the magnetic force. The current $I$ can be found using Ohm's law, $V = IR$, where $V$ is the voltage and $R$ is the resistance. The voltage is generated by the motion of the bar in the magnetic field and can be calculated using $V = BLv$, where $v$ is the velocity of the bar. Substituting $V = BLv$ into $V = IR$ gives $BLv = IR$, and substituting this into $F = BIL$ gives $F = B^2L^2v/R$. Solving for $B$ gives $B = \sqrt{FR/(L^2v)}$. Substituting $F = 0.080 N$, $R = 0.10 Ω$, $L = 0.50 m$, and $v = 0.50 m/s$ gives $B = \sqrt{(0.080 N)(0.10 Ω)/(0.50 m)^2(0.50 m/s)} = 0.25 T$.

Question 9

A closed loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the maximum emf induced in the loop is 5.0 V, what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies?

Accepted Answer

The maximum emf (ε) is given by Faraday's law: ε = -dΦ/dt, where Φ is the magnetic flux. For a circular loop, Φ = B * A = B * π * r^2. Given ε = 5.0 V, r = 2.0 m, and A = π * (2.0)^2, solve for dB/dt: 5.0 = π * (2.0)^2 * dB/dt, giving dB/dt = 0.40 T/s.

Question 10

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equation B(t) = 5.0t T, where t is time in seconds. If the induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, find the radius of the solenoid.

Accepted Answer

We use Faraday's law of electromagnetic induction to find the relation between the magnetic field and the induced electric field:

∮E⋅dl=−d/dt(∫B⋅dA)

where E is the induced electric field, B is the magnetic field, and the integral is taken over a closed loop around the solenoid.

For the conditions given, we can simplify this to:

E(2πr) = -d/dt(Bπr^2)

where r is the radius of the solenoid.

Taking the time derivative of B(t) = 5.0t T, we get:

d/dt(Bπr^2) = 5πr^2

Substituting this into the previous equation and solving for r:

E(2πr) = -5πr^2

r = -E/10

Substituting the given value for E and taking the absolute value, we get:

r = 0.94 m

Therefore, the answer is choice C.

Question 11

The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10^-2m² and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?

Accepted Answer

The peak output voltage (emf) of an AC generator is given by the formula: E_peak = NABω, where N is the number of turns, A is the area of each turn, B is the magnetic field strength, and ω is the angular velocity in rad/s. First, convert the frequency to angular velocity: ω = 2πf = 2π(60 Hz). Then, calculate E_peak = 125 turns × 3.0 × 10^-2 m² × 0.12 T × 2π(60 Hz) ≈ 170 V.

Question 12

A coil of 160 turns and area 0.20 m² is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magnetic field changes uniformly from 0.40 T in the +x direction to 0.40 T in the -x direction in 2.0 s. If the resistance of the coil is 16 Ω, at what rate is power generated in the coil?

Accepted Answer

The change in magnetic flux induces an emf, calculated using Faraday's law. The change in flux is 2 * 0.40 T * 0.20 m² = 0.16 Wb. The induced emf is (160 turns * 0.16 Wb) / 2 s = 12.8 V. The power generated is (12.8 V)² / 16 Ω = 10.24 W, approximately 10 W.

Question 13

A ten-loop coil having an area of 0.23 m2 and a very large resistance is in a 0.047-T uniform magnetic field oriented so that the maximum flux goes through the coil. The coil is then rotated so that the flux through it goes to zero in 0.34 s. What is the magnitude of the average emf induced in the coil during the 0.34 s?

Accepted Answer

The answer of A ten-loop coil having an area of...

Question 14

A circular loop of radius 0.10 m is rotating in a uniform external magnetic field of 0.20 T. Find the magnetic flux through the loop due to the external field when the plane of the loop and the magnetic field vector are
(a) parallel.
(b) perpendicular.
(c) at an angle of 30° with each other.

Accepted Answer

The answer of A circular loop of radius 0.10 m...

Question 15

A 200-loop coil of cross sectional area 8.5 cm² lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. (a) What is the magnitude of the change in the external magnetic flux enclosed by the coil? (b) What is the magnitude of the average voltage induced in the coil as the external flux is changing? (c) If the coil has a resistance of 4.0 ohms, what is the magnitude of the average current in the coil?

Accepted Answer

The answer of A 200-loop coil of cross sectional area...

Question 16

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.50e-20t T. What is the magnitude of the current induced in the coil at the time t = 2.0 s?

Accepted Answer

The induced emf (ε) is given by Faraday's law: ε = -dΦ/dt, where Φ = B*A is the magnetic flux. The area A = πr² = π(0.05)² m². The rate of change of B is dB/dt = -10e^(-20t) T/s. At t = 2.0 s, dB/dt = -10e^(-40) T/s. The induced emf is ε = A * |dB/dt|. The current I = ε/R = (A * |dB/dt|) / 0.20 Ω. Calculating gives approximately 2.6 mA.

Question 17

A 2.0-m long conducting wire is formed into a square and placed in the horizontal XY-Plane A uniform magnetic field is oriented 30.0° above the horizontal with a strength of 9.0 T. What is the magnetic flux through the square?

Accepted Answer

The side of the square is 0.5 m (since 2.0 m is the perimeter). The area is 0.25 m². The flux is B * A * cos(θ) = 9.0 T * 0.25 m² * cos(30°) = 1.95 T ∙ m², approximately 1.9 T ∙ m².

Question 18

As shown in the figure, a wire and a 10-Ω resistor are used to form a circuit in the shape of a square, 20 cm by 20 cm. A uniform but nonsteady magnetic field is directed into the plane of the circuit. The magnitude of the magnetic field is decreased from 1.50 T to 0.50 T in a time interval of 63 ms. The average induced current and its direction through the resistor, in this time interval, are closest to

Accepted Answer

The answer of As shown in the figure, a wire...

Question 19

Wire is wound on a square frame, 30 cm by 30 cm, to form a coil of 7 turns. The frame is mounted on a horizontal shaft through its center (perpendicular to the plane of the diagram), as shown in the figure. The coil is in clockwise rotation, with a period of 0.060 s. A uniform, horizontal, magnetic field of magnitude 0.40 T is present. At a given instant, the plane of the coil forms a 60° angle with the horizontal, as shown. At that instant, what is the magnitude of the emf induced in the coil?

Accepted Answer

The answer of Wire is wound on a square frame,...

Quiz 28: Electromagnetic Induction

A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?

A uniform magnetic field is applied perpendicular to the plane of a 60-turn circular coil with a radius of 6.0 cm and a resistance of 0.60 Ω. If the magnetic field increases uniformly from 0.20 T to 1.8 T in 0.20 s, what is the magnitude of the emf induced in the coil?

You are designing a generator to have a maximum emf of 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030 m2, what should be the frequency of the generator in a uniform magnetic field of 0.030 T?

Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm2, how many turns do you need?

The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10-2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?

A circular loop of radius 0.10 m is rotating in a uniform external magnetic field of 0.20 T. Find the magnetic flux through the loop due to the external field when the plane of the loop and the magnetic field vector are (a) parallel. (b) perpendicular. (c) at an angle of 30° with each other.

A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.50e-20t T. What is the magnitude of the current induced in the coil at the time t = 2.0 s?

A 2.0-m long conducting wire is formed into a square and placed in the horizontal XY-Plane A uniform magnetic field is oriented 30.0° above the horizontal with a strength of 9.0 T. What is the magnetic flux through the square?

You are designing a generator to have a maximum emf of 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030 m², what should be the frequency of the generator in a uniform magnetic field of 0.030 T?

Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm², how many turns do you need?

The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10^-2m² and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?