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 .
Answer:
.................... protons :)
Answer:
During a typical school day all forms of eneergy is being utilised and also transfer of energy takes place from one form to another.
Explanation:
Chemical energy- A bunsen burner burning a beaker filled with water.
Heat energy- The water in the beaker absorbing the heat from the burner.
Electrical energy- Running Fans and lights in a classroom by switches.
Solar energy- Solar energy harnessed by solar panels to run the fans and lights by converting it into electrical energy.
Potential energy- A ball being held by a student at a certain height possesses energy due to gravity.
Kinetic energy- The same ball being left by the boy from a certain height produces kinetic energy
The force on the proton is 17.4 N.
<h3>What is the force on the proton?</h3>
Now we know that the proton is positively charged and that the force on the charge as it moved through the magnetic field could be given by the relation; F = qvB
Where;
F = force
q = charge
v = velocity
B = magnetic field
Having said this, we can see that;
q = 1.601019 As or C
v = 2.4105 m/s
T = 4.5 T
F = 1.601019 As * 2.4105 m/s * 4.5 T
F = 17.4 N
Learn more about magnetic force:brainly.com/question/12824331
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Answer:
f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz
Explanation:
lets assume speed of sound is 350 m/s.
frequencies of a standing wave modes of an open-close tube of length L
fm = m(v/4L)
where m is 1,3,5,7......
and fm = mf1
where f1 = fundamental frequency
so therefore: f1 = 350 x 4 / 1.5
f1 = 58.3Hz
f2 = 3 x 58.3
f2 = 175Hz
f3 = 5 x 58.3
f3 = 291.6Hz
