Chapter 20, Problem 49P

To determine

The current carried by the solenoid.

Answer to Problem 49P

The solenoid should carry 94.3 A of current if it has 975 turns of the wire.

Explanation of Solution

Given:

The length of the solenoid, L = 30.0 cm = 0.3 m.

The magnetic field produced, B = 0.385 T.

Number of turns in coil, N = 975.

Formula used:

The magnetic field at the centre of the solenoid is given by $B={\mu }_0\left(\frac{N}{L}\right)I$ , where, I is the current and µ₀ is permittivity of free space.

  $$

Calculation: 

  $\begin{array}{l}B={\mu }_0\left(\frac{N}{L}\right)I\\ \therefore I=\frac{BL}{{\mu }_{0}N}=\frac{0.385\times 0.3}{4\pi \times {10}^{-7}\times 975}=94.3\text{A}\end{array}$

Conclusion:

Therefore, the current is 94.3 A.