Chapter 20, Problem 65P

To determine

Total magnetic field B inside the solenoid.

Answer to Problem 65P

  $B=40.50\text{T}$

Explanation of Solution

Given:

Loops of wire per meter in solenoid is 430.

Current carried is 25 A.

Permeability is 3000µ₀

Formula used:

The magnetic field is,

  $B=\frac{{\mu }_{0}NI}{L}$

Calculation:

Value of ${\mu }_0=4\pi \times {10}^{-7}{\text{N/m}}^{\text{2}}$

Here permeability is given 3000 µ₀

From $B=\frac{{\mu }_{0}NI}{L}$

  $B=\frac{3000\times 4\times 3.14\times {10}^{-7}\times 430\times 25}{1}$

  $B=40.50\text{T}$