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3 years ago

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Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in th

e same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 210? 10 da lo A. f/4 B. f/2 C. 3f/2 D. 2f

1 answer:

Alenkinab [10]3 years ago

3 0

Answer:

The option (B) is correct.

Explanation:

The magnetic force between the two current carrying wires is given by

$F =\frac{\mu o}{4\pi}\times \frac{2 I I'}{r}$

where, I and I' be the currents in the wires and r is the distance between the two wires.

Here, we observe that the force between the wires is inversely proportional to the distance between them.

So, when the distance is doubled , let the new force is F'.

$\frac{F'}{F}=\frac{r}{r'}\\\\\frac{F'}{F}=\frac{d}{2d}\\\\F'=\frac{F}{2}$

So, option (B) is correct.

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2 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

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(a) 69.3 J

The work done by the applied force is given by:

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where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

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The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

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3 years ago

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Answer:

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Explanation:

We know,

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Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

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