Answer:
0.021 V
Explanation:
The average induced emf (E) can be calculated usgin the Faraday's Law:
<u>Where:</u>
<em>N = is the number of turns = 1 </em>
<em>ΔΦ = ΔB*A </em>
<em>Δt = is the time = 0.3 s </em>
<em>A = is the loop of wire area = πr² = πd²/4 </em>
<em>ΔB: is the magnetic field = (0 - 1.04) T </em>
Hence the average induced emf is:
Therefore, the average induced emf is 0.021 V.
I hope it helps you!
Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
Answer:
λ = 2.7608 x 10⁻⁷ m = 276.08 nm
Explanation:
The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:
Work Function = hc/λ
where,
Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = longest wavelength capable of releasing electron.
Therefore,
7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)
<u>λ = 2.7608 x 10⁻⁷ m = 276.08 nm</u>
