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Aleksandr-060686
3 months ago
11

Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth

er two, which carries a current 0.5 I, but in the direction opposite from the first two wires. Two more wires with 0.5 I, in the same direct as the third wire, are placed the same distance as the third wire but on either side of the first two wires. In which direction are the net forces on the outer wires?

Physics
2 answers:
Sav [3.1K]3 months ago
8 0

Answer:

[ Find the attached file ]

Initially, I create a diagram illustrating the scenario. I considered the attractive force as positive and the repulsive force as negative. Afterward, I calculate the net force acting on the outer left wire caused by the other wires; the result is negative, indicating the presence of a repulsive force. Thus, the force is directed away from the wire, as depicted in the diagram.

Maru [3.3K]3 months ago
4 0
The force exerted is repulsive, meaning it moves away from the wire. To analyze this, one should create a diagram illustrating the scenario. If we define the attractive force as positive and the repulsive force as negative, computing the net force acting on the outer left wire from the other wires results in a negative value, indicating that the force is indeed repulsive, hence it points away from the wire as depicted in the accompanying figure.
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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
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Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Charge (Q₁) = 572.8 * 10^{-9} C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2} = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

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Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, F_1 = 32\ N (to the right)

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Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

F=F_1+F_2

F=32+(-16)

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

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