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A magnetic domain is a group of atoms aligns with magnetic poles. Domains are usually <span>light and dark stripes visible within each grain.</span>
The spring has been extended for 3.5 m
<u>Explanation:</u>
We have the formula,
PE =1/2 K X²
Rewrite the equation as
PE=1/2 K d²
multiply both the sides by 2/K to simplify the equation
2/k . PE= 1/2 K d² . 2/K
√d²=√2PE/K
Cancelling the root value and now we have,
d=√2PE/k
d=√2×98 J / 16N/m
d=√12.25
d=3.5 m
The spring has been extended for 3.5 m
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Answer:
(a): 
.
(b): 20.94 Volts.
Explanation:
<u>Given:</u>
- Area of the coil = A.
- Number of turns of the coil = N.
- Magnetic field in which the coil is placed = B.
- The frequency with which the coil is rotating = f.
<h2>
(a):</h2>
The magnetic flux linked with a coil is defined as
where,
is the area vector of the coil, directed along the normal to the plane of the coil.
= angle between the magnetic field and the area vector of the coil.
Assuming that the magnetic field is along the normal to the plane of the coil, initially.
Therefore, at any later time t, the angle which the magnetic field makes with the normal to the plane of the coil is is given by
Therefore, the magnetic flux linked with the coil at any time t is given by
According to the Faraday's law of electromagnetic induction, the emf induced in the coil is given by
<h2>
(b):</h2>
The amplitude of the alternating voltage is the maximum value of the induced emf in the coil, the induced emf in the coil is maximum when 
.
Therefore, the amplitude of the alternating voltage is given by
Given values are:
- N = 100 turns.
- A =
. - B = 0.1 T.
- f = 2000 rev/min =
Putting all these values,
Answer:
E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)
Explanation:
In this problem the electric field of a line of charge is requested, the expression for the electric field is
E = k ∫ dq / r²
where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)
let's use the definite linear density
λ₀ = dq / dx
dq = λ₀ dx
we replace and integrate
E = k λ₀ ∫ dx / x²
E = k λ₀ ( -1 / x)
we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞
E = - k λ₀ (1 /∞ - 1 / x₀)
E = k λ₀ / x₀
as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)
