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MariettaO [177]
3 years ago
10

How much current is in a circuit that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms? A. 0.44 amps B. 36 a

mps C. 2.3 amps D. 13 amps
Physics
1 answer:
bija089 [108]3 years ago
8 0

Answer:

C. 2.3A

Explanation:

V/Ω=A

9V / 4Ω = 2.25 ≅ 2.3A

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Two parallel wires carry equal currents in the opposite directions. Point A is midway between the wires, and B is an equal dista
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Answer:

Complete Question:(missing part)

The current in the left wire having magnitude is  = 210 A

The Current in the right wire having magnitude is = 10 A

Answer:

a.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )

Ratio = B 1/B 2 = 210/10 = 21

b.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1}}{3}-I_{1}} )

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Explanation:

Ratio of the magnitude of the magnetic field at point A to that at point B =?

The current in the left wire having magnitude is  = 210 A

The Current in the right wire having magnitude is = 10 A

distance between two wires = d

Given that both wires are carrying equal current but in opposite direction therefore

Magnetic field linked with the left wire at point A distance =d/2= according to biot savarts law

B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{d}{2} }

Magnetic field linked with the right wire at point B distance =d/2

B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{d}{2} }

Net magnetic field added

B net = B 1 + B 2

B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )

Ratio = B 1/B 2 = 210/10 = 21

Magnetic field linked to left wire at point B(d+ d/2)

B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{3d}{2} }

Magnetic field linked to right wire at point B(d+ d/2)

B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{3d}{2} }

Net magnetic field subtracted

B net = B 1 - B 2

B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1} }{3}-1 }  )

Ratio = B 1/B 2 = 210/3 X10 = 7

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