Drawing Conclusions Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magneti
2 answers:
Answer: No. Circuit should be closed.
Explanation:
A changing magnetic field induces emf in a circuit which is placed in it. This induced emf causes current to flow through it. This is given by Faraday's law which states that induced emf results due to rate of change of magnetic flux.
Current flows only through a closed circuit. Thus, when a wire is moved up and down in a magnetic field, it will not induce emf because the wire does not form a loop. It is an open circuit.
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Answer:
Velocity of the electron in the beam.
Radius of the circulating electrons due to the magnetic field.
Explanation:
We have a Mathematical expression for the force on a moving charge in a magnetic field as:
...........................(1)
where:
q= charge on the particle in coulomb
v= velocity of the charge projected into the magnetic field
B= intensity of the magnetic field in tesla
= angle between the velocity and direction of magnetic field
For the forces on rotating mass we have the formula:
..........................................(2)
where:
m= mass of the charged particle
v= velocity of projection of charge into the magnetic field
r= radius of the path traced by the charge in the magnetic field
From eq. (1) and (2) we can calculate the magnetic field .
Now,
Using Ampere's Law we have:
where:
I= current in the wire
The permeability of free space.
r= radial distance from the current carrying wire( in this case it is same as the radius of the circular path)
Answer:
he mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.
Explanation:
A radioactive material is transformed into another material by the emission of some particular radioactive ones, the most common being alpha and beta rays, which is why in the transformation process a certain amount of mass is lost. The process is described by the expression
N = No
From this expression the quantity half life time () is defined with time so that half of the atoms have been transformed
T_{1/2} = ln 2 /λ
in this case it does not indicate that T_{1/2}= 20 days is worth, for which periods have passed, in the first the number of radioactive atoms was reduced to half the number, leaving N´ and the second halved the number of nuclei that they were radioactive, leaving radioactive nuclei
first time of life
N´ = ½ N
second time of life
N´´ = ½ N´
N´´ = ¼ N
consequently in the sample at the end of these two decay periods we have, assuming that after each emission the atom is stable (non-radioactive). After the first emission there are n₁ = N / 2 stable atoms, after the second emission n₂ = ¼ N stable atoms are added and there are still n₃ = ¼ N radioactive atoms, so the total number of atoms is
n_total = n₁ + n₂ + n₃
Recall that the mass of the initial radioactive atoms is m₁, when transforming its mass of stable atoms is m₂ where
m₂ < m₁
therefore mass of
m_total = m₂ N / 2 + m₂ N / 4 + m₁ N / 4
m_total = m₂ ¾ N + m₁ ¼ N
m_total = N ( ¾ m₂ + ¼ m₁)
Since the mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.