Ask question

Ask question

All categories

2 years ago

7

In a model generator, a 310-turn rectangular coil 0.075 m by 0.18 m rotates with an angular frequency of 12.2 rad/s in a uniform

magnetic field of 0.64 T. What is the maximum emf induced in the coil

1 answer:

Novosadov [1.4K]2 years ago

4 0

This question involves the concepts of the induced emf, magnetic field, and angular frequency.

The maximum emf induced in the coil is "32.68 volts".

The following formula can be used to find out the induced emf in the coil.

$E =N\omega AB\ Sin(\omega t)$

But, for maximum induced emf sin function should have a value of 1. Therefore,

$E = N\omega AB$

where,

E = induced emf = ?

N = No. of turns in the coil = 310 turns

ω = angular frequency = 12.2 rad/s

A = Area of coil = (0.075 m)(0.18 m) = 0.0135 m²

B = magnetic field = 0.64 T

Therefore,

$E=(310)(12.2\ rad/s)(0.0135\ m^2)(0.64\ T)$

<u>E = 32.68 volts</u>

Learn more about induced emf here:

brainly.com/question/18441805?referrer=searchResults

You might be interested in

One of the foci for the moon's orbit would be the

suter [353]

The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.

5 0

2 years ago

Read 2 more answers

A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,

vfiekz [6]

The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
$J=F \Delta t$
But the impulse is also equal to the change in momentum of the cannonball:
$J=\Delta p$
If we put the two equations together, we find
$F \Delta t= \Delta p$
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
$\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s$

7 0

3 years ago

The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use

saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

$M = \frac{Nl}{f_ef_0}$

Where

N = Near point

l = distance between the object lens and eye lens

$f_0$ = Focal length

$f_e$ = Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

$M = \frac{25*10}{3*5}$

$M = 16.67\approx 17\\$

Therefore the correct answer is C.

3 0

3 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

2 years ago

The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7

stich3 [128]

Answer:

Electric field acting on the electron is 127500 N/C.

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Initial speed of electron, u = 0

Final speed of electron, $v=3\times 10^7\ m/s$

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}$

$a=2.25\times 10^{16}\ m/s^2$

According to Newton's law, force acting on the electron is given by :

F = ma

$F=9.1\times 10^{-31}\times 2.25\times 10^{16}$

$F=2.04\times 10^{-14}\ N$

Electric force is given by :

F = q E, E = electric field

$E=\dfrac{F}{q}$

$E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}$

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0

3 years ago