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gayaneshka [121]
3 years ago
9

A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section.

Physics
1 answer:
Mice21 [21]3 years ago
3 0

Answer:

Explanation:

We shall solve this question with the help of Ampere's circuital law.

Ampere's ,law

∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire

we shall find magnetic field at distance x . current enclosed in the area of circle of radius x

=  I x π x²  / π R²

= I x²  /  R²

B x 2π x = μ₀  x current enclosed

B x 2π x = μ₀  x  I x²  /  R²

B =  μ₀   I x  / 2π R²

Maximum magnetic B₀ field  will be when x = R

B₀ = μ₀I   / 2π R

Given

B = B₀ / 3

μ₀   I x  / 2π R² = μ₀I   / 2π R x 3

x = R / 3

b ) The largest value of magnetic field is on the surface of wire

B₀ = μ₀I   / 2π R

At distance x outside , let magnetic field be B

Applying Ampere's circuital law

∫ B dl = μ₀ I

B x 2π x = μ₀ I

B = μ₀ I / 2π x

Given B = B₀ / 3

μ₀ I / 2π x = μ₀I   / 2π R x 3

x = 3R .

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Answer: 13.1 μH

Explanation:

Given

length of heating coil, l = 1 m

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No of loops, N = 400

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μ = 4π*10^-7 = 1.26*10^-6 T

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L = [1.26*10^-6 * 400 * 400* 6.5*10^-5] / 1

L = 1.26*10^-6 * 1.6*10^5 * 6.5*10^-5

L = 1.31*10^-5

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Thus, from the calculations above, we can say that the total self inductance of the solenoid is 13.1 μH

7 0
3 years ago
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Use the equation for motion to answer the question.
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I choose the option D.
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X = 2 + 15 x 1 + 0 = 17 m
8 0
3 years ago
The tire on this drag racer is severely twisted: The force of the road on the tire is quite large(most likely several times the
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When the force on the tire is equal to the weight of the car, the car is reaching a stability as a result of increase in motion.

But when the force of the load on the tire is quite large(most likely several times the weight of the car) and is directed forward, then, the car is at high speed.

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A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration
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Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

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Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

b)

acceleration , a = -1.5 j m/s^2

Using first equation of motion

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v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

5 0
3 years ago
A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri
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The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
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Solving for the static friction force (F), 
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Therefore, the maximum force of static friction is 1.794 N. 



5 0
3 years ago
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