A solenoid of 1.5m length and 4.0cm diameter possesses 100 turns per meter. A current of 5amperes is flowing through it. The magnetic induction at axis inside the solenoid is:
A. $2\pi \times \,{10^{ - 4}}{\rm{T}}$
B. $2\pi \times \,{10^{ - 5}}{\rm{T}}$
C. $2\pi \times \,{10^{ - 2}}{\rm{gauss}}$
D. $2\pi \times \,{10^{ - 5}}{\rm{gauss}}$
Answer
611.4k+ views
Hint: A solenoid is a long coil of wire wrapped in many turns in a cylinder. When a current passes through it, it creates a nearly uniform magnetic field inside. The direction of the magnetic field depends on the direction of the current flowing.
The magnetic induction at axis inside the solenoid is calculated using the formula,
${\rm{B}}\; = {\mu _0}{\rm{nI}}$
Where B is the magnetic field density, ${\mu _0}$ is the permeability of free space, n is the number of turns and ${\rm{I}}$ is the current flowing through the coil. The magnetic field generated along the axis of a solenoid is proportional to the number of turns per unit length and the current owing in it.
Complete step-by-step answer:
Given,
Length of the solenoid,$l\; = \;1.5{\rm{m }}$
The diameter of the solenoid, ${\rm{d}}\;{\rm{ = }}\;{\rm{4}}{\rm{.0}}\;{\rm{cm}}$
Number of turns per cm, n $ = 100\;{\text{turns per cm }}$
${\mu _0}\, = \;4\pi \times \;{10^{ - 7}}$
Current ${\rm{I}}\;{\rm{ = }}\;{\rm{5A}}$
The formula for the magnetic field at axis inside the solenoid is given as,
${\rm{B}}\; = {\mu _0}{\rm{nI}}$
This formula is also known as Ampere’s Law.
$ \Rightarrow \;{\rm{B}}\,{\rm{ = }}4\pi \times \;{10^{ - 7}} \times \;100 \times 5 = \,2\pi \times {10^{ - 4}}\,{\rm{T}}$
Hence the correct option is (A).
Note: We have to apply the concept of magnetic field production when a current flows through a current-carrying conductor or coil. The direction of the magnetic field depends on the direction of the current, and the value of the magnetic field can be calculated from the number of turns in the coil, the permeability of free space and the amount of current owing through the coil. A solenoid is used as an electromagnet which can provide a simple and strong magnetic field. It is mainly used to convert electrical motion to mechanical motion like in plungers.
The magnetic induction at axis inside the solenoid is calculated using the formula,
${\rm{B}}\; = {\mu _0}{\rm{nI}}$
Where B is the magnetic field density, ${\mu _0}$ is the permeability of free space, n is the number of turns and ${\rm{I}}$ is the current flowing through the coil. The magnetic field generated along the axis of a solenoid is proportional to the number of turns per unit length and the current owing in it.
Complete step-by-step answer:
Given,
Length of the solenoid,$l\; = \;1.5{\rm{m }}$
The diameter of the solenoid, ${\rm{d}}\;{\rm{ = }}\;{\rm{4}}{\rm{.0}}\;{\rm{cm}}$
Number of turns per cm, n $ = 100\;{\text{turns per cm }}$
${\mu _0}\, = \;4\pi \times \;{10^{ - 7}}$
Current ${\rm{I}}\;{\rm{ = }}\;{\rm{5A}}$
The formula for the magnetic field at axis inside the solenoid is given as,
${\rm{B}}\; = {\mu _0}{\rm{nI}}$
This formula is also known as Ampere’s Law.
$ \Rightarrow \;{\rm{B}}\,{\rm{ = }}4\pi \times \;{10^{ - 7}} \times \;100 \times 5 = \,2\pi \times {10^{ - 4}}\,{\rm{T}}$
Hence the correct option is (A).
Note: We have to apply the concept of magnetic field production when a current flows through a current-carrying conductor or coil. The direction of the magnetic field depends on the direction of the current, and the value of the magnetic field can be calculated from the number of turns in the coil, the permeability of free space and the amount of current owing through the coil. A solenoid is used as an electromagnet which can provide a simple and strong magnetic field. It is mainly used to convert electrical motion to mechanical motion like in plungers.
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