Answer:
Explanation:
Given that, .
Frequency
f = 60Hz
Number of turns
N = 125turns
Surface area of coil
A = 3 × 10^-2 m²
Magnetic field
B = 0.12T
Voltage peak to peak? I.e the EMF
EMF is given as
ε = —dΦ/dt
Where Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Where N is number of turns
B is magnetic field
A is the cross sectional area
And θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, your magnetic flux becomes
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
I.e Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
The peak to peak voltage is 169.65 V
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Answer:
The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.
Explanation:
Given that,
Wavelength = 617 nm
Width of slit
Distance between the slit and screen L= 2.83 m
Third dark fringe m = 3
We need to calculate the distance on the screen from the center of the central bright fringe to the third dark fringe on either side
Using formula of distance
Put the value into the formula
Hence, The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm