Answer:
The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Explanation:
The magnetic field B at the center of the solenoid is given by
B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.
So B = μ₀ni
= 4π × 10⁻⁷H/m × 1100 × 1.5 A
= 4π × 10⁻⁷H/m × 1650 A-turns/m
= 20734.5 × 10⁻⁷T
= 2.07345 × 10⁻³ T
≅ 2.1 × 10⁻³ T
So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Angle of reflection = <span>θ [ As it is equal to angle of incidence ]
Direction of reflected ray is always opposite to direction of incident ray, i.e., away from the surface of mirror.
Hope this helps!</span>
Answer:
216 units
Explanation:
F1 = kQ1Q2/r^2 = 36 (Coulomb's law)
F2 = k(2Q1)(3Q2)/r^2
= 6[kQ1Q2/r^2]
= 6F1
= 6 x 36
= 216 units
Please write in English because I do not understand
Answer:
that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
Explanation:
Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.
Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.
In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.
If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.