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

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SUMMARY

The discussion focuses on calculating the charge, current, and electromotive force (EMF) in a circular coil subjected to a reversing magnetic field. A 100-turn coil with a diameter of 2 cm and a resistance of 50 Ohms is analyzed under a uniform magnetic field of 1 T. The total charge passing through the coil is calculated to be 0.001256 C, the average current is determined to be 0.01256 A, and the average EMF is found to be 0.626 V. The calculations are confirmed as correct by other participants in the discussion.

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  • Familiarity with the formula for calculating EMF (E = IR)
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Seiya
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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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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.
 

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