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# A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns if it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid is:A. $6 \cdot 024 \times {10^3}T$B. $8 \cdot 024 \times {10^3}T$C. $5 \cdot 024 \times {10^3}T$D. $7 \cdot 024 \times {10^3}T$

Last updated date: 13th Jun 2024
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Hint:-A solenoid is a coil of wire made as an electromagnet. A solenoid’s main purpose is to generate a controlled magnetic field by tightening wires to form a helical coil. Arrangements of wires to form coil in helical shape is known as a solenoid.
Formula used: The formula of the magnetic field of the solenoid coil is given by,
$B = {\mu _o}nI$
Where B is the magnetic field I is the current in the circuit and n is the number of terms per meter of the length.

Complete step-by-step solution
It is given that the length of solenoid is 0.6 m and the radius of the wire is 2 cm the number of turns of the solenoid is 600 and if the current carried by the coil is 4 A then we need to find the value of the magnetic field that has been produced here by the solenoid.
Let us calculate the number of turns per meter of the length of the solenoid i.e. the value of n.
$\Rightarrow n = \dfrac{{600}}{{0 \cdot 6}}$
$\Rightarrow n = 1000\dfrac{1}{m}$
The formula of the magnetic field is given by,
$B = {\mu _o}nI$
Where B is the magnetic field I is the current in the circuit and n is the number of terms per meter of the length.
Replace the value of current in the coil and the value of n is the above equation.
$\Rightarrow B = {\mu _o}nI$
Replace $n = 1000\dfrac{1}{m}$ and$I = 4A$.
$\Rightarrow B = {\mu _o}nI$
$\Rightarrow B = {\mu _o}\left( {1000} \right)\left( 4 \right)$
$\Rightarrow B = \left( {4\pi \times {{10}^{ - 7}}} \right) \times \left( {1000} \right) \times \left( 4 \right)$
$\Rightarrow B = 5 \cdot 024 \times {10^3}T$
The magnetic field of the solenoid is equal to$B = 5 \cdot 024 \times {10^3}T$. The correct answer for this problem is option C.

Note:- A solenoid can generate a uniform magnetic field in the space when current passes through the wire. The length of any solenoid is always greater than the diameter of the solenoid because this helps us to bring a greater uniformity in the magnetic field.