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riadik2000 [5.3K]
3 years ago
7

A parallel-plate capacitor with circular plates of radius R is being charged by a battery, which provides a constant current. At

what radius inside and outside the capacitor gap is the magnitude of the induced magnetic field equal to half of its maximum value? (Use any variable or symbol stated above as necessary.)
Physics
1 answer:
NikAS [45]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the magnetic field.

According to the information, the magnetic field INSIDE the plates is,

B=\frac{1}{2} \mu \epsilon_0 r

Where,

\mu =Permeability constant

\epsilon_0 =Electromotive force

r = Radius

From this deduction we can verify that the distance is proportional to the field

B \propto r

Then the distance relationship would be given by

\frac{r}{R} = \frac{B}{B_{max}}

r =\frac{B}{B_{max}} R

r = \frac{0.5B_{max}}{B_{max}}R

r = 0.5R

On the outside, however, it is defined by

B = \frac{\mu_0 i_d}{2\pi r}

Here the magnetic field is inversely proportional to the distance, that is

B \not\propto r

Then,

\frac{r}{R} = \frac{B_{max}{B}}

r = \frac{B_{max}{B}}R

r = \frac{B_{max}{0.5B_{max}}}R

r = 2R

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Answer:

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In this problem, we have

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f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz

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<span>
The force on m is:</span>

<span>
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= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

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