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Nana76 [90]
2 years ago
9

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

ough a loop of length a = 0.27 m and width b = 0.63 m, placed a distance c = 0.10 m from the sheet? (Enter the magnitude.)
Physics
1 answer:
oksian1 [2.3K]2 years ago
6 0

Answer:

\phi _{B} =0.855 T-m^{-2}

Explanation:

given data

density of current sheet = 0.40 A/m

length a = 0.27 m

width b = 0.63 m

For infinite sheet, magnetic field is given as

B = \mu _{O}J

magnetic flux is given as

\phi _{B} = BA

                   = \mu _{O}Jab

                   = 4\pi *0.40*0.27*0.63

\phi _{B} =0.855 T-m^{-2}

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Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp
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Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is 1.6\times10^{-19}C. Then the total charge that is flowing from B to A is:

1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

A + B = 3.68 (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

A + 2.416 = B - 2.416

-A + B = 2.416 + 2.416 = 4.832 (2)

Add equation (1) to equation (2) we have

2B = 3.68 + 4.832 = 8.512

B = 8.512 / 2 = 4.256 \mu C

A = 3.68 - B = 3.68 - 4.256 = -0.576 \mu C

6 0
2 years ago
As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
Andreyy89

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

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2 years ago
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The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
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Answer:

Explanation:

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Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
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Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

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The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

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Charge on each plate = Surface charge density on each plate x area of each plate

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Answer:

Hello there Dude answer is B :D hope it helped mark me brainliest.

8 0
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