magnetic field produced by A solenoid carrying conductor explain it
Answers
Magnetic Field Produced by a Current-Carrying Solenoid
A solenoid is a long coil of wire (with many turns or loops, as opposed to a flat loop). Because of its shape, the field inside a solenoid can be very uniform, and also very strong. The field just outside the coils is nearly zero. This figure shows how the field looks and how its direction is given by RHR-2.
A diagram of a solenoid. The current runs up from the battery on the left side and spirals around with the solenoid wire such that the current runs upward in the front sections of the solenoid and then down the back. An illustration of the right hand rule 2 shows the thumb pointing up in the direction of the current and the fingers curling around in the direction of the magnetic field. A length wise cutaway of the solenoid shows magnetic field lines densely packed and running from the south pole to the north pole, through the solenoid. Lines outside the solenoid are spaced much farther apart and run from the north pole out around the solenoid to the south pole.
(a) Because of its shape, the field inside a solenoid of length
l
is remarkably uniform in magnitude and direction, as indicated by the straight and uniformly spaced field lines. The field outside the coils is nearly zero. (b) This cutaway shows the magnetic field generated by the current in the solenoid.
The magnetic field inside of a current-carrying solenoid is very uniform in direction and magnitude. Only near the ends does it begin to weaken and change direction. The field outside has similar complexities to flat loops and bar magnets, but the magnetic field strength inside a solenoid is simply
B
=
μ
0
nI
(
inside a solenoid
)
,
where
n
is the number of loops per unit length of the solenoid
(
n
=
N
/
l
, with
N
being the number of loops and
l
the length). Note that
B
is the field strength anywhere in the uniform region of the interior and not just at the center. Large uniform fields spread over a large volume are possible with solenoids, as the example below implies.
Example: Calculating Field Strength inside a Solenoid
What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current?
Strategy
To find the field strength inside a solenoid, we use
B
=
μ
0
nI
. First, we note the number of loops per unit length is
n
=
N
l
=
2000
2.00
m
=
1000
m
−
1
=
10
cm
−
1
.
Solution
Substituting known values gives
B
=
μ
0
nI
=
(
4
π
×
10
−
7
T
⋅
m/A
)
(
1000
m
−
1
)
(
1600 A
)
=
2
.01 T.
Discussion
This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000 loops squeezed into a meter’s length would produce significant heating. Higher currents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields.