The field strength needed to produce a 24.0 V peak emf is 0.73T.
To find the answer, we need to know about the expression of emf.
What's the expression of peak emf produced in a rotating rectangular loops?
- The peak emf produced in a rotating loops= N×B×A×w
- N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
- So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
- N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
- Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
- Now, B= 24/(300×0.00261×42)
B= 24/(300×0.00261×42) = 0.73T
Thus, we can conclude that the magnetic field is 0.73T.
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Answer:
Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinations of equipment would be needed to bring one liter of water to 85°C? a. ... Various pieces of safety equipment are used in the lab to provide protection against injury.
Explanation:
Answer:
no where is the main part of the question dude
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Answer:
The tunnel probability for 0.5 nm and 1.00 nm are 
and 
respectively.
Explanation:
Given that,
Energy E = 2 eV
Barrier V₀= 5.0 eV
Width = 1.00 nm
We need to calculate the value of 
Using formula of
Put the value into the formula
(a). We need to calculate the tunnel probability for width 0.5 nm
Using formula of tunnel barrier
Put the value into the formula
(b). We need to calculate the tunnel probability for width 1.00 nm
Hence, The tunnel probability for 0.5 nm and 1.00 nm are and respectively.
