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anastassius [24]
3 years ago
6

Spinning a loop of wire between the poles of a magnet will induce an electric current. What condition would minimize the induced

current?
A. The plane of the loop is perpendicular to the magnetic field.

B. The magnetic flux through the wire is minimized.

C. The plane of the loop is parallel to the magnetic field.

D. The plane of the loop is halfway between the parallel and perpendicular.
Physics
1 answer:
DochEvi [55]3 years ago
8 0
B. The other choices all happen at some point in the rotation but you want to maximize the flux and then rotate it as fast as possible
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Suppose that a circular parallel-plate capacitor has radius R0 = 3.0 cm and plate separation d = 5.0 mm. A sinusoidal potential
Arturiano [62]

Answer:

B=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}cos(\omega t)

Explanation:

By the information of the statement you have that the sinusoidal potential difference is given by:

V=V_osin(\omega t)=V_osin(2\pi ft)=150sin(2\pi (60)t)    (1)

In order to calculate the induced magnetic field in between the plates, you first take into account the following formula, which is the Ampere-Maxwell law:

\int B\cdot ds=\mu_o \epsilon_o\frac{d\Phi_E}{dt}+\mu_oI_c           (2)

B: induced magnetic field

μo: magnetic permeability of vacuum = 4π*10^-7 A/T

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Ic: conduction current

ФE: electric flux

There is no conduction current in between the plates, then Ic = 0A

Next, you calculate dФE/dt, as follow:

The electric field, and the electric flux, are:

E=\frac{V}{d}=\frac{V_osin(\omega t)}{d}\\\\\Phi_E=EA

d: separation between plates = 5.0mm = 5.0*10^-3 m

A: area of the circular plates = \pi R_o^2

Ro: radius of the circular capacitor = 3.0cm = 0.03m

Thus, dФE/dt is:

\frac{d\Phi_E}{dt}=\frac{d(EA)}{dt}=\pi R_o^2\frac{d}{dt}[\frac{V_osin(\omega t)}{d}]\\\\\frac{d\Phi_E}{dt}=\frac{\pi \omega R_o^2 V_o}{d}cos(\omega t)       (3)

The induced magnetic field is calculated by taking into account that the integral of the equation (2) is:

\int B \cdot ds=B\int ds=B(2\pi r)         (4)

Next, you replace the results of (3) and (4) into the equation (2) and you solve for B:

B(2\pi r)=\mu_o \epsilon_o (\frac{\pi \omega R_o^2 V_o}{d}cos(\omega t))\\\\B=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}cos(\omega t)   (5)

The last expression is de induced magnetic field in between the plates in terms of t and r

Another way of expressing the  formula (5) is as follow:

B=B_ocos(\omega t)\\\\B_o=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}

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Hope that helps!
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