Question

A current carrying toroid winding is internally filled with lithium having susceptibility \( \chi = 2.1 \times 10^{-5} \). What is the percentage increase in the magnetic field in the presence of lithium over that without it?

Hint:

The magnetic field in the presence of a material increases in proportion to the magnetic susceptibility of the material.

Answer 1

The magnetic field in a material with magnetic susceptibility \( \chi \) is given by:
\[ B = \mu_0 H (1 + \chi) \] where \( \mu_0 \) is the permeability of free space, and \( H \) is the magnetic field intensity. The percentage increase in the magnetic field due to the presence of lithium is:
\[ {Percentage increase} = \frac{B_{{with lithium}} - B_{{without lithium}}}{B_{{without lithium}}} \times 100 = \frac{\mu_0 H (\chi)}{\mu_0 H} \times 100 \] Substituting the value of \( \chi \):
\[ {Percentage increase} = \chi \times 100 = 2.1 \times 10^{-5} \times 100 = 0.0021\% \]