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kondaur [170]
3 years ago
7

You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Whic

h of the following increases the energy density? (Select all that apply.)
a. Increasing only the length of the solenold while keeping the turns per unit lengh flxed
b. increasing the number of turns per unit length on the solenold
c. increasing the cross-sectional area of the solenoid
d. none of these
e. increasing the current in the solenoid
Physics
1 answer:
Alinara [238K]3 years ago
5 0

Answer:

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

Explanation:

As we know that energy density depends on the strength of the magnetic field. The magnetic field strength depends on the no of turns of the solenoid and the current passing through it. The greater the number of turns per unit length, greater the current passing through it, more stronger the magnetic field is. As

B = μ₀nI

n = no of turns

I = current through the wire

So the right options are

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

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While studying physics at the library late one night, you noticethe image of the desk lamp reflected from the varnished tabletop
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<h2>Answer: 1.73</h2>

Explanation:

The situation described here is known as polarization by reflection. This was discovered by Scottish physicist David Brewster and then formulated the law that bears his name:

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This polarization happens when the light incides at a specific angle, called the Brewster angle (\theta_{B}), which is given by the following formula (<u>taking into account that generally the magnetic permeabilities of the two media involved do not vary):</u>

tan\theta_{B}=\frac{n_{2}}{n_{1}}     (1)

Where n_{2} is the index of refraction of the second medium (the varnish in this case) and n_{1}=1 is the index of refraction of the first medium (the air).

Now, if we are told the angle between the incident and reflected rays is 120\°, this means the incident angle is the half (60\°), which is the Brewster angle in this case.

So, \theta_{B}=60\°   (2)

Rewriting (1) with this incident ray angle:

tan(60\°)=\frac{n_{2}}{1}   (3)

n_{2}=tan(60\°)  

Finally we obtain the index ofrefraction of the varnish:

n_{2}=1.732

5 0
3 years ago
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