Calculating Charge, Current, and EMF in a Reversing Magnetic Field"

In summary, the coil has a diameter of 2 cm and a resistance of 50 Ohms, and is perpendicular to a uniform magnetic field of magnitude 1 T. After the field is suddenly reversed, the total charge that passes through the coil is 0.001256 C, the average current in the coil is 0.01256 A, and the average emf in the coil is 0.626 V.
  • #1
Seiya
43
1
Hey lads, if anyone can, please check If I am correct... Thanks :)

A 100 turn circular coil has a diameter of 2 cm and a resistance of 50 Ohms. The plane of the coil is perpendicular to a uniform magnetic field of magnitude 1 T. The direction of the field is suddenly reversed.

(a) Find the total charge that passes through the coil. If the reversal takes 0.1 s, find
(b) the average current in the coil and
(c) the average emf in the coil


So I got...

I= dq/dt

Е=-dф/dt

Edt/R=dQ

Integral of dQ = Q = -1/R (int from ф0 to фf) dф

= -1/R(фf-ф0) = -(delta)ф/R

ф0 (field goes inside, n vector goes inside) = NBAcos() = +NBA
фf (field goes outside, n vector goes inside) = NBAcos() = -NBA

So Q = - (-NBA-(NBA))/R = 2NBA/R = .001256C

(b) I = dQ/dt and dt= .1s
= .01256A (since dQ = Q because Q0= 0 )

(c) E=IR = 0.626V

Am I correct? If not can you please point out any mistakes I have made... thanks i very much appreciate it =)
 
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  • #2
Seiya said:
Hey lads, if anyone can, please check If I am correct... Thanks :)

A 100 turn circular coil has a diameter of 2 cm and a resistance of 50 Ohms. The plane of the coil is perpendicular to a uniform magnetic field of magnitude 1 T. The direction of the field is suddenly reversed.

(a) Find the total charge that passes through the coil. If the reversal takes 0.1 s, find
(b) the average current in the coil and
(c) the average emf in the coil


So I got...

I= dq/dt

Е=-dф/dt

Edt/R=dQ

Integral of dQ = Q = -1/R (int from ф0 to фf) dф

= -1/R(фf-ф0) = -(delta)ф/R

ф0 (field goes inside, n vector goes inside) = NBAcos() = +NBA
фf (field goes outside, n vector goes inside) = NBAcos() = -NBA

So Q = - (-NBA-(NBA))/R = 2NBA/R = .001256C

(b) I = dQ/dt and dt= .1s
= .01256A (since dQ = Q because Q0= 0 )

(c) E=IR = 0.626V

Am I correct? If not can you please point out any mistakes I have made... thanks i very much appreciate it =)

This looks completely right to me.
 
  • #3


Hello! Your calculations seem to be correct. However, I would like to clarify a few points. Firstly, in part (b), you have correctly calculated the average current to be 0.01256A, but this is only the average current during the reversal period of 0.1s. If you want to find the overall average current in the coil, you would need to take into account the initial current before the reversal as well.

Secondly, in part (c), you have correctly calculated the average emf to be 0.626V. However, this is the average emf during the reversal period of 0.1s. If you want to find the overall average emf in the coil, you would need to take into account the initial emf before the reversal as well.

Overall, your calculations seem to be correct and you have a good understanding of the concepts involved. Keep up the good work!
 

Related to Calculating Charge, Current, and EMF in a Reversing Magnetic Field"

1. What is a motional EMF?

A motional EMF (electromotive force) is an induced voltage or potential difference that is created in a conductor as it moves through a magnetic field. This phenomenon is based on Faraday's law of electromagnetic induction.

2. How is a motional EMF produced?

A motional EMF is produced when a conductor moves through a magnetic field or when there is a change in the strength of the magnetic field that the conductor is moving through. This movement creates a changing magnetic flux, which induces an electric field and subsequently an EMF in the conductor.

3. What is the relationship between motional EMF and velocity?

The magnitude of the motional EMF is directly proportional to the velocity of the conductor and the strength of the magnetic field. This means that a faster-moving conductor or a stronger magnetic field will result in a larger induced EMF.

4. Can a motional EMF be measured?

Yes, a motional EMF can be measured using a voltmeter. The voltage measured will depend on the velocity of the conductor, the strength of the magnetic field, and the length of the conductor within the magnetic field.

5. What are some real-world applications of motional EMF?

Motional EMF has various applications in our daily lives, such as in generators and motors. It is also used in devices like magnetic levitation trains, which use motional EMF to propel themselves forward. Additionally, motional EMF is used in induction cooktops, where the changing magnetic field induces currents in the cookware, generating heat for cooking.

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