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*Question:* A long solenoid with an air core has turns per meter
and a cross-sectional area of
. The current flowing
around
the solenoid increases from 0 to in .
A plane loop of wire consisting of turns, which is of cross-sectional area
and resistance
, is placed around the
solenoid close to its centre. The loop is orientated such that
it lies in the plane perpendicular to the axis
of the solenoid.
What is the magnitude of the emf
induced
in the coil? What current does does this emf drive around the coil?
Does this current circulate in the same direction as the current flowing in
the solenoid, or in the opposite direction?

*Answer:* We must, first of all, calculate the magnetic
flux linking the coil. The magnetic field is confined to the region
inside the solenoid (the field generated outside a *long* solenoid
is essentially negligible). The magnetic field runs along the axis of the
solenoid, so it is directed perpendicular to the plane of the coil. Thus,
the magnetic flux linking a single turn of the coil is the product of the
area of the magnetic-field-containing region and the strength
of the perpendicular field. Note that, in this case, the magnetic flux does
not depend on the area of the coil, as long as the magnetic-field-containing
region lies completely within the coil.
The magnetic flux linking the
whole coil is the flux linking a single turn times the number of
turns in the coil. Thus,

Now, the magnitude of the magnetic field generated by the solenoid
is given by (see Sect. 8.8)

so the magnetic flux linking the coil can be written

This magnetic flux increases because the current flowing in the
solenoid increases.
Thus, the time rate of change of the magnetic flux is given by

By Faraday's induction law, the emf generated around the coil is

Ohm's law gives

as the current induced in the coil.
According to Lenz's law, the current induced in the coil is such as to
oppose the increase in the magnetic flux linking the coil. Thus, the current
in the coil must circulate in the *opposite* direction to the current in
the solenoid, so that the magnetic field generated by the the former current, in the middle of
the coil, is oppositely directed to that generated by the latter current.
The fact that the current in the above formula is *negative*
is indicative of the fact that
this current
runs in the opposite direction to the current flowing around the solenoid.

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** Up:** Magnetic Induction
** Previous:** Example 9.1: Faraday's law
Richard Fitzpatrick
2007-07-14