Answer:
V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)
Explanation:
Attached is the full solution
Answer:
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Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
Answer:
The smallest film thickness is 117 nm.
Explanation:
Light interference on thin films can be constructive or destructive. Constructive interference is dependent on the film thickness and the refractive index of the medium.
For the first interference (surface nearest to viewer), the minimum thickness can be expressed as:
where n is the refractive index of the bubble film.
Therefore,
∴ 
