Question 1

What is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 A? (μ₀ = 4π × 10^-7 T • m/A)

Accepted Answer

The energy density $u$ in a magnetic field is given by the formula $u = \frac{B^2}{2\mu_0}$, where $B$ is the magnetic field strength and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}$). The magnetic field $B$ around a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$, where $I$ is the current and $r$ is the distance from the wire. Substituting $I = 12 \, \text{A}$ and $r = 0.25 \, \text{m}$, and then calculating $u$, results in the energy density value corresponding to choice B.

Question 2

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 answer of A closed loop conductor that forms a...

Question 3

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 4

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 length of the square is 0.5 m (since 2.0 m / 4 sides = 0.5 m). The area of the square is 0.5 m * 0.5 m = 0.25 m². The magnetic flux is given by Φ = B * A * cos(θ), where B = 9.0 T, A = 0.25 m², and θ = 30°. Thus, Φ = 9.0 T * 0.25 m² * cos(30°) = 1.1 T • m².

Question 5

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?

Accepted Answer

The answer of The coil in a 60-Hz ac generator...

Question 6

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 7

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 8

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 9

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 maximum emf (E_max) generated by a coil in a magnetic field can be calculated using the formula: E_max = NABωsin(θ), where N is the number of turns, A is the area, B is the magnetic field strength, ω is the angular frequency (ω = 2πf, where f is the frequency), and θ is the angle between the magnetic field and the normal to the coil. Since the maximum emf occurs when sin(θ) = 1, the equation simplifies to E_max = NABω. Solving for frequency (f), we get f = E_max / (2πNAB).Given: E_max = 8.0 V, N = 200 turns, A = 0.030 m^2, B = 0.030 T, we find:f = 8.0 / (2π * 200 * 0.030 * 0.030) = 7.1 Hz.

Question 10

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A.If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s,how fast is the current changing?

Accepted Answer

The answer of At a certain instant the current flowing...

Question 11

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,30...

Question 12

An insulated wire of diameter 1.0 mm and negligible resistance is wrapped tightly around a cylindrical core of radius 5.0 cm and length 30 cm to build a solenoid. What is the energy stored in this solenoid when a current I = 0.20 A flows through it? (μ₀ = 4π × 10^-7 T • m/A)

Accepted Answer

The answer of An insulated wire of diameter 1.0 mm...

Question 13

How much energy is stored in a room 3.0 m by 4.0 m by 2.4 m due to the earth's magnetic field with a strength of 5.0 × 10^-5 T? (μ₀ = 4π × 10^-7 T • m/A)

Accepted Answer

The energy stored in a magnetic field can be calculated using the equation:

E = (1/2) μ V B^2

Where E is energy, μ is the permeability of the material (in this case, vacuum), V is the volume, and B is the magnetic field strength.

Substituting the given values, we get:

E = (1/2) (4π × 10^-7 T•m/A) (3.0 m × 4.0 m × 2.4 m) (5.0 × 10^-5 T)^2
E ≈ 29 mJ

Therefore, the correct answer is B.

Question 14

A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it.What is the winding density of the solenoid? (μ₀ = 4π × 10^-7 T • m/A)

Accepted Answer

The answer of A solenoid of length 0.700 m having...

Question 15

What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10^-4 m²? (μ₀ = 4π × 10^-7 T • m/A)

Accepted Answer

We can use the formula for self-inductance of a solenoid:
L = μ * n^2 * A * l / 2
where μ is the permeability of free space, n is the number of turns per unit length (in this case, n = N/L, where N is the total number of turns and L is the length of the solenoid), A is the cross-sectional area, and l is the length of the solenoid.

Plugging in the given values, we get:
n = 100 turns / 0.30 m = 333.3 turns/m
L = (4π × 10^-7 T • m/A) * (333.3 turns/m)^2 * (1.00 × 10^-4 m^2) * (0.30 m) / 2 
L = 4.19 µH

Therefore, the best choice is C, 4.19 µH.

Question 16

At what rate would the current in a 100-mH inductor have to change to induce an emf of 1000 V in the inductor?

Accepted Answer

The induced emf (voltage) in an inductor is given by the formula $V = L \frac{dI}{dt}$, where $V$ is the voltage in volts, $L$ is the inductance in henrys, and $\frac{dI}{dt}$ is the rate of change of current in amperes per second. Rearranging the formula to solve for $\frac{dI}{dt}$, we get $\frac{dI}{dt} = \frac{V}{L}$. Substituting the given values, $V = 1000$ V and $L = 100$ mH = 0.1 H, we find $\frac{dI}{dt} = \frac{1000}{0.1} = 10,000$ A/s.

Question 17

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

The magnetic field strength within the solenoid is given by 
B(t) = 5.0t T
We need to find the radius of the solenoid.

At a distance of 2.0 m from the axis of the solenoid, the induced electric field is given as 
E = 1.1 V/m

Using the formula for magnetic field due to a solenoid with circular cross-section at the center,
B = μ₀nI

Where,
μ₀ - magnetic constant = 4π×10⁻⁷ Tm/A
n - number of turns per unit length of solenoid
I - current passing through the solenoid

At the center of the solenoid, the magnetic field is uniform and is given by
B = μ₀nI

The induced electric field at a distance r from the axis of the solenoid is given by 
E = -dB/dt

Since we have B(t), we can find dB/dt which is 
dB/dt = 5.0 T/s

At a distance of r = 2.0 m, the induced electric field is given as 1.1 V/m.

We can equate both the expressions for E and find nI which is 
nI = E/μ₀dB/dt = 1.1/(4π×10⁻⁷×5.0) = 4.395×10⁶ A

The number of turns per unit length of the solenoid(n) can be found as  
n = 4.395×10⁶ turns/m

The radius of the solenoid (R) is given by 
R = √(n/I)/π

Substituting the values, we get 
R = √(4.395×10⁶/5)/π = 0.94 m

Hence, the answer is (C) 0.94 m.

Question 18

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...

Quiz 29: Changing Magnetic Fields

What is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 A? (μ0 = 4π × 10-7 T • m/A)

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?

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?

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 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?

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?

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A.If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s,how fast is the current changing?

An insulated wire of diameter 1.0 mm and negligible resistance is wrapped tightly around a cylindrical core of radius 5.0 cm and length 30 cm to build a solenoid. What is the energy stored in this solenoid when a current I = 0.20 A flows through it? (μ0 = 4π × 10-7 T • m/A)

How much energy is stored in a room 3.0 m by 4.0 m by 2.4 m due to the earth's magnetic field with a strength of 5.0 × 10-5 T? (μ0 = 4π × 10-7 T • m/A)

A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it.What is the winding density of the solenoid? (μ0 = 4π × 10-7 T • m/A)

What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ0 = 4π × 10-7 T • m/A)