Power = I^2 x R
Energy = Power x Time
Answer:
All the given options will result in an induced emf in the loop.
Explanation:
The induced emf in a conductor is directly proportional to the rate of change of flux.
where;
A is the area of the loop
B is the strength of the magnetic field
θ is the angle between the loop and the magnetic field
<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.
<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.
Option C has a similar effect with option A, thus both will result in an induced emf.
Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.
Therefore, all the given options will result in an induced emf in the loop.
Explanation:
The charge on the electron is,
The electric field at a distance r from the electron is :
Where
k is the electrostatic constant,
We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.
So, the correct option is " the electric field is directed toward the electron and has a magnitude of . Hence, this is the required solution.
Answer:
The separation between the charges was decreased by a factor of 0.2
Explanation:
The Coulomb's force between two charges is given by;
r₂ = 0.2r₁
Therefore, the separation between the charges was decreased by a factor of 0.2.
Answer:
(A) L = 115.3kgm²/s
(B) dL/dt = 94.1kgm²/s²
Explanation:
The magnitude of the angular momentum of the rock is given by the foemula
L = mvrSinθ
We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.
Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =
115.3 kgm²/s
(B) The magnitude of the rate of angular change in momentum is given by
dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²