A solenoid of radius 10 cm carrying current 0.29 A and having total 200 turns. If magnetic field inside solenoid is\( 2.…

We are given:

• Radius of solenoid = \(r=10\)cm (not needed in calculations)
• Current = \(I=0.29\) A
• Total number of turns = \(N=200\)
• Magnetic field inside the solenoid = \(B=2.9\times 1{0}^{−4}\)

We need to determine the length of the solenoid (\(L\)).

The magnetic field inside a solenoid is given by:

\(B={\mu }_{0}nI\)

where:

• \({\mu }_{0}\)​ is the permeability of free space, \({\mu }_{0}=4\pi \times 1{0}^{−7}\)

• \(n\)n is the number of turns per unit length, given by:

  \(n=\frac{N}{L}\)​

• \(I\) is the current.

Rewriting the equation:

\(B={\mu }_{0}\frac{N}{L}I\)

Solving for \(L\):

\(L=\frac{{\mu }_{0}NI}{B}\)​

\(L=\frac{(4\pi \times 1{0}^{−7})(200)(0.29)}{2.9\times 1{0}^{−4}}\)

​ \(L=\frac{(4\pi \times 1{0}^{−7}\times 200\times 0.29)}{2.9\times 1{0}^{−4}}\)

\(L=8\pi \text{ cm}\)

\(B\ 8\pi \text{ cm}\)