Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d
1 answer:
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 × m)
d = (3.30 × 563 × ) ÷ (0.0047)
= 1.8579 × ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
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Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.