A rectangular loop of wire that can rotate about an axis through its center is placed between the poles of a magnet in a magneti
c field with a strength of 0.40 T. The length of the loop L is 0.16 m and its width w is 0.040 m. What is the magnetic flux through the loop when the plane of the loop is perpendicular to the magnetic field?
1 answer:
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Answer:
The first part can be solved via conservation of energy.
For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.
where because we are looking for the case where the car loses contact.
Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.
Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.
Answer:
hello your question has some missing values attached below is the complete question with the missing values
answer :
a) 0.083 secs
b) 0.33 secs
c) 3e^-4/3
Explanation:
Given that
g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft
initial displacement = 1 ft above equilibrium
mass = weight / g = 4/32 = 1/8
damping force = instanteous velocity hence β = 1
a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>
time mass passes through equilibrium = 1/12 seconds = 0.083
<u>b) Calculate the time at which the mass attains its extreme displacement </u>
time when mass attains extreme displacement = 1/3 seconds = 0.33 secs
<u>c) What is the position of the mass at this instant</u>
position = 3e^-4/3
attached below is the detailed solution to the given problem
