Ask question

Ask question

All categories

4 years ago

10

A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, w

ith the plane of the coil parallel to the field. In 0.230 ss , it is rotated so that the plane of the coil is perpendicular to the field.
(a) What is the change in the magnetic flux through the coil due to the rotation?
(b) What is the magnitude of the average emf induced in the coil during the rotation?
(c) What is the average current induced in the coil during the rotation?

1 answer:

Travka [436]4 years ago

7 0

Answer:

(a) ΔФ = -0.109W

(b) emf = 28.43V

(c) Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

$\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)$ (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

$\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W$

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

$emf=-N\frac{\Delta \Phi}{\Delta t}$ (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

You replace the values of the parameters in the equation (2):

$emf=-(60)(\frac{-0.109W}{0.230s})=28.43V$

The induced emf is 28.43V

(c) The induced current in the coil is given by:

$I_{in}=\frac{emf}{R}$ (3)

R: resistance of the coil (it is necessary to have this value)

emf :induced emf = 28.43V

You might be interested in

Which form of energy is associated with an object’s position?

dybincka [34]

Answer:

potential

Explanation:

4 0

3 years ago

Read 2 more answers

Pressure that increases with depth in a swimming pool is called ______________ pressure.

lukranit [14]

Hydrostatic pressure.

4 0

4 years ago

Estimate under what set of conditions it is possible, despite the spherical shape of the working electrode, to use linear diffus

goldfiish [28.3K]

When the experimental measurements diverge at very low values of the voltages. When the ratio of current to saturated current is very small. That is, i/I less than or equal to 1/100. When there is a finite current at Voltage

V = 0

Diffusion will have influence on current - voltage curve in the working electrode which is ionization detector.

The estimation under set of conditions that is possible, despite the spherical shape of the working electrode, to use linear diffusion equations to theoretically predict the current vs voltage curve expected in this experiment will be:

When the experimental measurements diverge at very low values of the voltages
when the ratio of current to saturated current is very small. That is, i/I less than or equal to 1/100
When there is a finite current at Voltage V = 0

Where

i = electric current

I = saturated current

V = voltage supplied

The curve expected will therefore give exponential curve from positive to negative domain because of the diffusion of ions from different directions

Learn more here: brainly.com/question/23985719

8 0

3 years ago

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s$

D) 5.6 rad

The angular displacement of the wheel is given by the equation

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where we have

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel

$\omega_f = 4.00 rad/s$ is the final angular velocity

$\alpha=+1.67 rad/s^2$ is the angular acceleration

Solving for $\theta$ ,

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad$

3 0

3 years ago

Runs 4 m North and then runs 9 m West. Calculate Distance and Displacement.

mixas84 [53]

Answer:

Take north as y and west as -x.

distance=√(x^2+y^2)

displacement= tan theta(tan^-1) (y/x)

answers should be, (9.85m, 66.04°)

8 0

3 years ago