Question

A solenoid of 1.5 metre length and 4.0cm diameter possesses $10$ turn per cm. A current of 5 ampere is flowing through it. The magnetic induction at axis inside the solenoid is $({\mu _0} = 4\pi \times {10^{ - 7}}weberam{p^{ - 1}}{m^{ - 1}})$
A) $2\pi \times {10^{ - 3}}tesla$
B) $2\pi \times {10^{ - 5}}tesla$
C) $2\pi \times {10^{ - 2}}gauss$
D) $2\pi \times {10^{ - 5}}gauss$

Answer 1

Recall that the solenoid is a type of electromagnet. It is in the form which is in the form of a cylinder. When an electric current is passed through a solenoid then it produces a magnetic field. Solenoids are basically made up of ferromagnetic materials.

Complete step by step solution:
Step I:
Given that the length of the solenoid is $l = 1.5m$
$d = 10turns$
Current flowing is $I = 5A$

Step II:
According to Ampere’s Circuital law, the magnetic field produced by a current carrying conductor around a closed path is directly proportional to the length of the number of turns of the solenoid and the current flowing through it.
Let B is the strength of the magnetic field around the solenoid then the formula can be written as
$B = {\mu _0}nI$
where ${\mu _0}$is the magnetic constant

Step III:
Substituting all the given values in the formula, and solving
$B = 4\pi \times {10^{ - 7}} \times 10 \times 5$
$B = 2\pi \times {10^{ - 5}}T$

Therefore, the magnetic induction inside the solenoid is $2\pi \times {10^{ - 5}}T$.
$\therefore$ Option (B) is the correct answer.

Note: It is important to remember that the magnetic field around a coiled wire will be much stronger than the magnetic field around a straight wire. The strength of the magnetic field can be increased if the number of coils in the solenoid is increased. This is because each turn in the wire will contribute and has its own magnetic field, hence increasing the total magnetic field. Also if an iron core is added in the solenoid, then the field strength increases. The magnetic field at the centre is stronger than at the corners in a solenoid.