
A solenoid is $2.0$m long and $3.0$cm in diameter. It has $5$ layers of winding $1000$ turns each and carries a current of $5.0$A. What is the magnetic field at its center?
Answer
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Hint: Solenoid provides a means to produce a strong magnetic field. The solenoid is a coil of wire which acts as an electromagnet when a flow of current passes through it.
The solenoid is a long cylindrical helix. It consists of a large number of turns of insulated copper wire winded over it to the tube. When an electric current is passed through the solenoid, a magnetic field is produced within and around the solenoid.
Formula Used:
We will be using the formula of ampere’s law to calculate the strength of the magnetic field. It is given by,
$B = {\mu _0} \times n \times I$
Where $I$- Current through wire
$n$-Number of coils per meter
$B$-magnetic field
${\mu _0}$-permeability
To find the magnetic field at the center of the solenoid.
Complete step by step solution:
In a solenoid, the magnetic field lines inside the solenoid are nearly parallel which shows that the magnetic field is uniform and parallel to the axis of the solenoid. It has $5$ layers of winding $1000$ turns each.
Given: Length of the solenoid $L = 2m$
Diameter $d = 3cm$
Number of turns per meter $ = \dfrac{{5 \times 1000}}{2}$
$n = 2500{m^{ - 1}}$
The magnetic field within the solenoid depends upon the current and number of turns. By using the formula,
$B = {\mu _0} \times n \times I$
Current $I = 5A$
$\therefore B = 4\pi \times {10^{ - 7}} \times 2500 \times 5$
$ \Rightarrow B = 1.57 \times {10^{ - 2}}T$
Note: Solenoids can convert electric current to mechanical action and force. The Earth’s magnetic field is about half gauss. The magnetic field is the strongest inside the coil.
A solenoid is a temporary magnet and it can be demagnetized by stopping the current through the solenoid. Magnetic field lines cannot like the electric field lines (converge or diverge). So there is no magnetic field outside the solenoid. Hence the density of the magnetic field lines outside is reduced.
The solenoid is a long cylindrical helix. It consists of a large number of turns of insulated copper wire winded over it to the tube. When an electric current is passed through the solenoid, a magnetic field is produced within and around the solenoid.
Formula Used:
We will be using the formula of ampere’s law to calculate the strength of the magnetic field. It is given by,
$B = {\mu _0} \times n \times I$
Where $I$- Current through wire
$n$-Number of coils per meter
$B$-magnetic field
${\mu _0}$-permeability
To find the magnetic field at the center of the solenoid.
Complete step by step solution:
In a solenoid, the magnetic field lines inside the solenoid are nearly parallel which shows that the magnetic field is uniform and parallel to the axis of the solenoid. It has $5$ layers of winding $1000$ turns each.
Given: Length of the solenoid $L = 2m$
Diameter $d = 3cm$
Number of turns per meter $ = \dfrac{{5 \times 1000}}{2}$
$n = 2500{m^{ - 1}}$
The magnetic field within the solenoid depends upon the current and number of turns. By using the formula,
$B = {\mu _0} \times n \times I$
Current $I = 5A$
$\therefore B = 4\pi \times {10^{ - 7}} \times 2500 \times 5$
$ \Rightarrow B = 1.57 \times {10^{ - 2}}T$
Note: Solenoids can convert electric current to mechanical action and force. The Earth’s magnetic field is about half gauss. The magnetic field is the strongest inside the coil.
A solenoid is a temporary magnet and it can be demagnetized by stopping the current through the solenoid. Magnetic field lines cannot like the electric field lines (converge or diverge). So there is no magnetic field outside the solenoid. Hence the density of the magnetic field lines outside is reduced.
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