Average induced emf in coil with turns

In summary, the conversation discusses calculating the average emf induced in a rectangular coil with 95.0 turns, dimensions of 29.0 cm by 40.0 cm, and a rotation of 39.0 degrees from a position parallel to a magnetic field of 1.60 T to a position perpendicular to the field in 0.120 seconds. The equation used is ε=ΔB/Δt*(A*cos 39°), which results in an average emf of 1.20. However, this is incorrect as it does not take into account the factor N and the initial and final magnetic flux values.
  • #1
susan14
2
0

Homework Statement


A closely wound rectangular coil of 95.0 turns has dimensions of 29.0 cm by 40.0 cm. The plane of the coil is rotated from a position where it makes an angle of 39.0 degrees with a magnetic field of 1.60 T to a position perpendicular to the field. The rotation takes 0.120 s.
What is the average emf induced in the coil?


Homework Equations





The Attempt at a Solution


ε=ΔB/Δt*(A*cos 39°)
ε=1.6/.120*(.116*cos39)
=1.20
This is the wrong answer!
 
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  • #2
susan14 said:

Homework Statement


A closely wound rectangular coil of 95.0 turns has dimensions of 29.0 cm by 40.0 cm. The plane of the coil is rotated from a position where it makes an angle of 39.0 degrees with a magnetic field of 1.60 T to a position perpendicular to the field. The rotation takes 0.120 s.
What is the average emf induced in the coil?


Homework Equations





The Attempt at a Solution


ε=ΔB/Δt*(A*cos 39°)
ε=1.6/.120*(.116*cos39)
=1.20
This is the wrong answer!

Welcome to PF, susan14! :smile:


Which formula do you have for the emf?

I have:
$$\mathcal{E}=-N {d\Phi \over dt}$$
where either ##\Phi = B_{\textit{perpendicular to A}} A## or ##\Phi = B A_{\textit{perpendicular to B}}##.




The average ##\mathcal{E}## would be:
$$\mathcal{E}_{average}=-N {\Delta\Phi \over \Delta t}=-N {\Phi_t - \Phi_0 \over t - 0}$$
where ##\Phi_0## is the magnetic flux at time t=0, and ##\Phi_t## is the magnetic flux at the final time t.

What would you get for ##\Phi_0## and for ##\Phi_t##?
And what did you do with the factor N?
 

What is the formula for calculating the average induced emf in a coil with multiple turns?

The formula for calculating the average induced emf in a coil with turns is:
E = -N(dΦ/dt)
Where:
E = average induced emf
N = number of turns in the coil
dΦ/dt = rate of change of magnetic flux through the coil

How does the number of turns in a coil affect the average induced emf?

The number of turns in a coil has a direct relationship with the average induced emf. This means that as the number of turns increases, the average induced emf also increases. This is because a coil with more turns has a larger surface area for the magnetic flux to pass through, resulting in a higher induced emf.

What factors can affect the rate of change of magnetic flux and thus impact the average induced emf?

There are several factors that can affect the rate of change of magnetic flux and impact the average induced emf. These include:
- The strength of the magnetic field
- The speed at which the coil is moved relative to the magnetic field
- The angle between the coil and the magnetic field
- The shape and size of the coil

Can the average induced emf in a coil with turns be negative?

Yes, the average induced emf in a coil with turns can be negative. This occurs when the magnetic flux through the coil decreases, resulting in a negative rate of change of magnetic flux. This can happen when the coil is moved away from the magnetic field or when the magnetic field itself weakens.

How is the average induced emf in a coil with turns related to Faraday's Law of Induction?

The average induced emf in a coil with turns is directly related to Faraday's Law of Induction, which states that the magnitude of the induced emf is equal to the rate of change of magnetic flux through the coil. The more turns in the coil, the larger the surface area for the magnetic flux to pass through and the higher the average induced emf.

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