A 90 cm long solenoid has six layers of windings of 450 turns each. If the diameter of solenoid is 2.2 cm and current carried is 6 A, then the magnitude of magnetic field inside the solenoid, near its centre is:
A . 50πG
B . 60πG
C . 72πG
D . 80πG
Answer
619.2k+ views
Hint: Solenoid is a ring kind of structure having turns around the ring. You can assume bangles having turns around it. When current flows through it a magnetic field is generated and e.m.f is induced in it. To solve this question, use ampere's law. Number of turns is equal to ratio of turns and length. Field inside the solenoid depends on current and number of turns. S.I. The unit of magnetic field is Tesla (T) but it can be measured in gauss. Denoted by G. magnetic field depends on the number of lines passing through it.
Complete step-by-step answer:
Question explanation- We have a wire of 90cm long. Make circles of that wire. That circle has 6 layers of 450 turns to each. The diameter of a solenoid is 2.2cm and current is 6A.
Aim: Find magnitude of magnetic field inside the solenoid near its centre.
We know that the Maxwell equation has four equations. Use fourth equation i.e. Ampere's law.
Ampere law- It stated that ‘’ the line integral of magnetic field of induction vector B around any closed path in free space is equal to absolute permeability of free space times the total current flowing through are bound by the path”.
Ampere law is given by,
\[\oint{\overrightarrow{B.}\overrightarrow{dl}={{\mu }_{o}}I}\]
For fixed layers of winding the total number of turns \[=6\times 450=2700\] turns. Since we have six layers and 450 turns to each layer.
Now number of turn per unit length is given by,
\[n=\dfrac{N}{l}=\dfrac{2700}{90\times {{10}^{-2}}}=3000\]
The field inside the solenoid near the centre is given by,
$\begin{align}
& B={{\mu }_{0}}nI \\
& =4\pi \times {{10}^{-7}}\times 3000\times 6=72\pi G \\
\end{align}$
Then the magnitude of magnetic field inside the solenoid, near its centre is:$72\pi G$
Answer is (C)
Note: Magnetic field is corresponding to magnetic flux. Magnetic flux is nothing but the number of lines passing through per unit of area. Magnetic flux measure in weber. Tesla can be written as $Wb/{{m}^{2}}$. Solenoid is stronger than coil magnet and can obtain desired power of magnet. Magnet will show magnetism only when a current passes through solenoid/coil.
Complete step-by-step answer:
Question explanation- We have a wire of 90cm long. Make circles of that wire. That circle has 6 layers of 450 turns to each. The diameter of a solenoid is 2.2cm and current is 6A.
Aim: Find magnitude of magnetic field inside the solenoid near its centre.
We know that the Maxwell equation has four equations. Use fourth equation i.e. Ampere's law.
Ampere law- It stated that ‘’ the line integral of magnetic field of induction vector B around any closed path in free space is equal to absolute permeability of free space times the total current flowing through are bound by the path”.
Ampere law is given by,
\[\oint{\overrightarrow{B.}\overrightarrow{dl}={{\mu }_{o}}I}\]
For fixed layers of winding the total number of turns \[=6\times 450=2700\] turns. Since we have six layers and 450 turns to each layer.
Now number of turn per unit length is given by,
\[n=\dfrac{N}{l}=\dfrac{2700}{90\times {{10}^{-2}}}=3000\]
The field inside the solenoid near the centre is given by,
$\begin{align}
& B={{\mu }_{0}}nI \\
& =4\pi \times {{10}^{-7}}\times 3000\times 6=72\pi G \\
\end{align}$
Then the magnitude of magnetic field inside the solenoid, near its centre is:$72\pi G$
Answer is (C)
Note: Magnetic field is corresponding to magnetic flux. Magnetic flux is nothing but the number of lines passing through per unit of area. Magnetic flux measure in weber. Tesla can be written as $Wb/{{m}^{2}}$. Solenoid is stronger than coil magnet and can obtain desired power of magnet. Magnet will show magnetism only when a current passes through solenoid/coil.
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