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Free_Kalibri [48]
3 years ago
7

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 19.0 cmcm and carries

a clockwise current of 20.0 AA , as viewed from above, and the outer wire has a diameter of 38.0 cmcm .What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires? What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
Physics
1 answer:
Lemur [1.5K]3 years ago
7 0

Answer:

Explanation:

magnetic field due to circular wire

= μ₀ i / 2r

i is current and r is radius of coil .

Magnetic fields due to inner coil

μ₀ x 20 / (2 x 9.5 x 10⁻²)

Magnetic field due to outer coil

= μ₀ x I / (2 x 19 x 10⁻²) , I is the current to be calculated

Total field

μ₀ x 20 /( 2 x 9.5 x 10⁻²) +μ₀ x I / (2 x 19 x 10⁻²)  = 0

20 + I /2 = 0

I = - 40 A

Current required is 40 A , and it will be in opposite direction.

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2 years ago
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The period of a pendulum is given by
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
\frac{T_m}{T_e} =  \frac{2 \pi  \sqrt{ \frac{L}{g_m} } }{2 \pi  \sqrt{ \frac{L}{g_e} } }=    \sqrt{ \frac{g_e}{g_m} }
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is g_e = 9.81 m/s^2 while on the Moon is g_m=1.63 m/s^2, the ratio between the period on the Moon and on Earth is
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3 0
3 years ago
In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.
garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...

Find the equivalent resistance of the right branch of the circuit:

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\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411

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Mass is indirectly proportional to acceleration, so, lighter the object greater would be it's acceleration...

A) 0.10 kg is lightest among them, so it's your answer

6 0
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The answer is c. 
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