Answer:State the question so I can answer
Explanation:
Answer:
Explanation:
The angular position of first intensity maxima is given by the expression
= λ / d where λ is wave length of light and d is slit separation
putting the values given in the problem. ,
.02 = λ₁ / D
Now wavelength has been increased to 2λ₁
angular separation of first maxima
= 2λ₁ /D
= 2 x .02
= .04 radians.
Answer:
331.75 V
Explanation:
Given:
Number of turns of the coil, N = 40 turns
Area, A = 0.06 m²
Magnetic Field, B = 0.4 T
Frequency, f = 55 Hz
Maximum induce emf, E₀ = NABω
but ω = 2πf
Maximum induce emf, E₀ = NAB(2πf₀)
Maximum induce emf, E₀ = 2πNABf₀
Where;
N is number of turns of the coil
A is area
B is magnetic field
ω is the angular velocity
f is the frequency
E₀ = 2 × π × 40 × 0.06 × 0.4 × 55
E₀ = 342.81 V
The maximum induced emf is 331.75 V
Answer:
Explanation:
Let v be the velocity acquired by electron in electric field
V q = 1/2 m v²
V is potential difference applied on charge q , m is mass of charge , v is velocity acquired
2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²
v² = 844 x 10¹²
v = 29.05 x 10⁶ m /s
Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .
Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron
= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶
= 79.02 x 10⁻¹³ N .
Minimum force will be zero when electron moves along the direction of magnetic field .
