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
Zarrin [17]
3 years ago
9

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

tic field. A field of 1.5 T is directed along the positive z-direction, which is upward. Of the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process.
Physics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

You might be interested in
Mike rides his horse with a constant speed of 20 km/h. How far can he travel in 4 hours?
gizmo_the_mogwai [7]

Answer:

Mike can travel 80 Km in 4 hours

4 0
2 years ago
Two loudspeakers emit 600 Hz Hz notes. One speaker sits on the ground. The other speaker is in the back of a pickup truck. You h
Firdavs [7]

Answer:

The truck's speed is 4.04 m/s.

Explanation:

Given that,

Emit frequency = 600 Hz

Beat = 7.00 beat/sec

We need to calculate the truck's speed

Using formula of speed

\text{frequency observed}=\text{frequency emitted}\times\dfrac{v}{v+v_{source}}

Where, v = speed of sound

Put the value into the formula

(600-7)=600\times(\dfrac{343}{343-v_{truck}})

v_{truck}=\dfrac{600\times343-593\times343}{593}

v_{truck}=4.04\ m/s

Hence, The truck's speed is 4.04 m/s.

3 0
2 years ago
In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8
pashok25 [27]
Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

6 0
3 years ago
If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____
Tju [1.3M]
<span>Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span>           n </span>⇒ remaining mass of cobalt after 3 years
          T ⇒ decaying period
           t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
 3/t= 0.567

t = 3÷0.567
  = 5.290626524

the half-life of Cobalt-60 is 5.29 years. 

<span>           
</span><span>
</span>
8 0
2 years ago
a 0.4 kg block rests on a desk. the coefficient of static friction is 0.2. You push the side of the block but do not have a spri
ladessa [460]
Static friction is greater than Applied force
7 0
3 years ago
Read 2 more answers
Other questions:
  • In proof testing of circuit boards, the probability that any particular diode will fail is 0.01. Suppose a circuit board contain
    15·1 answer
  • In a wet cell battery what is the electrolyte? <br> copper<br><br> liquid acid<br><br> zinc
    13·2 answers
  • Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
    6·1 answer
  • Stu wanted to calculate the resistance of a light bulb connected to a 4.0-V battery, with a resulting current of 0.5 A. He used
    12·2 answers
  • How much work is done when you push a crate horizontally with 150 N across a 13-m floor
    7·1 answer
  • F(x)= 10x-5<br>What is the value of f-1(-4) ?​
    10·2 answers
  • You drop a rock off a bridge. When the rock has fallen 4 m, you drop a second rock. As the two rocks continue to fall, what happ
    10·1 answer
  • An object in a certain direction with an acceleration in the perpendicular direction
    14·1 answer
  • Hi, how do you draw two opposite rays and name them? Is it two of these "&lt;---&gt;" that intersect? Or two separate rays?
    12·1 answer
  • Which of these objects has the greatest inertia?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!