A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) = B0 cos(ωt) ˆz.
A circular loop of wire, of radius a/2 and resistance R, is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time
1 answer:
Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
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Answer:
<em>0.85c </em>
Explanation:
Rest mass of Kaon = 494 MeV/c²
Rest mass of proton = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy = 494c² MeV
for the proton, rest energy = 938c² MeV
Recall that the rest energy, and the total energy are related by..
= γ
which can be written in this case as
= γ
...... equ 1
where = total energy of the kaon, and
= rest energy of the kaon
γ = relativistic factor =
where
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
=
......equ 2
where is the total energy of the kaon, and
is the rest energy of the proton.
From =
= 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = = 1.89
1.89 = 1
squaring both sides, we get
3.57( 1 - ) = 1
3.57 - 3.57 = 1
2.57 = 3.57
= 2.57/3.57 = 0.72
= 0.85
but,
v/c = 0.85
v = <em>0.85c </em>