1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ivan
3 years ago
10

A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni

tude of the field is allowed to vary in time according to B = Bmax e-t/τ , where Bmax and τ are constants. The field has the constant value Bmax for t < 0. Find the emf induced in the loop as a function of time. (Use the following as necessary: A, Bmax, t, and τ.) g
Physics
1 answer:
mina [271]3 years ago
7 0

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

You might be interested in
What is the radius of the 5th orbital in hydrogen?
iren2701 [21]

Answer:

So, the radius of fifth Bohr orbital of hydrogen is 1. 3225 nm.

Explanation:

pls mark me brainless hope this helps loves x!

6 0
2 years ago
Read 2 more answers
A car starts from rest and accelerates uniformly over a time of 5.37 seconds for a distance of 101 m. Determine the acceleration
castortr0y [4]

Answer:

8.10m s-2

Explanation:

8 0
4 years ago
A +26.3 uC charge qy is repelled by a force
Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0
3 years ago
HURRY!!!!
VMariaS [17]

Answer:

Ted is correct

Explanation:

The equation for gravitational potential energy is PE = m·g·h

The equation for gravitational kinetic energy is KE = 1/2·m·v²

Where:

m = Mass of the object (The racing car)

g = Acceleration due to gravity

h = The height to which the object is raised

v = Velocity of motion of the object

From the principle of conservation of energy, energy can neither be created nor destroyed but changes from one form to another, we have;

Potential energy gained from location at height h = Kinetic energy gained as the object moves down the level ground

m·g·h = 1/2·m·v² canceling like terms gives

g·h = 1/2·v²

v = (√2·g·h)

If the speed is doubled, we have

2·v = 2× (√2·g·h) =  (√2·g·4·h)

Therefore, if 2·v = v₂ then v₂ =  (√2·g·4·h)

Since g, the acceleration due to gravity, is constant, it means that the initial  height must be multiplied or increased 4 times to get the new height, that is we have;

v₂ =  (√2·g·4·h) = (√2·g·h₂)

Where:

4·h = h₂

Which gives;

v₂² = 2·g·h₂

1/2·v₂² = g·h₂

1/2·m·v₂² = m·g·h₂ Just like in the first relation

Therefore, Ted is correct s they need to go up four times the initial height to double the speed.

5 0
3 years ago
The Jamaican bobsled team was moving at a velocity of 50 m/s, then they hit the brakes on their sled to decelerate at a uniform
atroni [7]

Answer:

The time it took the bobsled to come to rest is 10 s.

Explanation:

Given;

initial velocity of the bobsled, u = 50 m/s

deceleration of the bobsled, a = - 5 m/s²

distance traveled, s = 250 m

Apply the following kinematic equation to determine the time of motion of the bobsled;

s = ut + ¹/₂at²

250 = 50t + ¹/₂(-5)t²

250 = 50t - ⁵/₂t²

500 = 100t - 5t²

100 = 20t -t²

t² - 20t + 100 = 0

t² -10t - 10t + 100 = 0

t (t - 10) - 10(t - 10) = 0

(t - 10)(t - 10) = 0

t = 10 s

Therefore, the time it took the bobsled to come to rest is 10 s.

3 0
3 years ago
Other questions:
  • Based on ohms law, the unit of volts is equivalent to
    14·1 answer
  • What would happen if a proton were added to the nucleus of the atom? (Check all that apply)
    6·1 answer
  • A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the
    13·1 answer
  • A string is strung horizontally with a fixed tension. A wave of frequency 90 Hz is sent along the string, and it has a wave spee
    10·1 answer
  • A 440 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s. The rocket engine, when it is fired, exert
    11·1 answer
  • Given that a pipe having a diameter of 1.5 feet and a height of 10 feet, what is the psi at 5 feet ?
    15·1 answer
  • Temperature has a(n) _____ effect on the pressure and volume of a gas.
    9·1 answer
  • What is a quantum field?
    7·2 answers
  • Why solar panels are usually painted black​
    10·1 answer
  • Laptop computers are made with batteries, but they must also be plugged into outlets to charge the batteries. which is true rega
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!