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masha68 [24]
3 years ago
11

Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point

x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?
Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

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How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 23.0 cm in diameter to p
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Electric field is given by

E=\dfrac{kq}{r^2}\\\Rightarrow q=\dfrac{Er^2}{k}\\\Rightarrow q=\dfrac{1150\times 0.115^2}{8.99\times 10^9}\\\Rightarrow q=1.69174\times 10^{-9}\ C

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n=\dfrac{1.69174\times 10^{-9}}{1.6\times 10^{-19}}\\\Rightarrow n=10573375000

Number of excess electrons is 10573375000

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Answer:

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