Question

The free space inside a current carrying toroid is filled with a material of susceptibility \( \chi = 2 \times 10^{-2} \). The percentage increase in the value of magnetic field inside the toroid will be:

• A. 2%
• B. 0.2%
• C. 0.1%
• D. 1%

Hint:

For a material with susceptibility \( \chi \), the percentage increase in the magnetic field inside a toroid is equal to the susceptibility \( \chi \) expressed as a percentage.

Answer 1

The Correct Option is A

The magnetic field \( B \) inside a toroid depends on the permeability of the material filling the toroid. The relation between the permeability of the material and the permeability of free space is given by: \[ \mu = \mu_0 (1 + \chi) \] where: - \( \mu_0 \) is the permeability of free space, - \( \chi \) is the susceptibility of the material, - \( \mu \) is the permeability of the material inside the toroid. The magnetic field inside the toroid is given by: \[ B = \frac{\mu I}{2 \pi r} \] where: - \( I \) is the current, - \( r \) is the radius of the toroid, - \( \mu \) is the permeability of the material inside the toroid. Now, the percentage increase in the magnetic field when the toroid is filled with the material is given by: \[ {Percentage increase in } B = \frac{\Delta B}{B_{{initial}}} \times 100 \] The initial magnetic field \( B_{{initial}} \) is given by \( B_{{initial}} = \frac{\mu_0 I}{2 \pi r} \), and the final magnetic field \( B_{{final}} \) is given by \( B_{{final}} = \frac{\mu_0 (1 + \chi) I}{2 \pi r} \). Thus, the percentage increase in the magnetic field is: \[ \frac{\Delta B}{B_{{initial}}} = \frac{B_{{final}} - B_{{initial}}}{B_{{initial}}} = \frac{\frac{\mu_0 (1 + \chi) I}{2 \pi r} - \frac{\mu_0 I}{2 \pi r}}{\frac{\mu_0 I}{2 \pi r}} = \chi \] Substituting \( \chi = 2 \times 10^{-2} \): \[ {Percentage increase in } B = 2 \times 10^{-2} \times 100 = 2\% \] Thus, the percentage increase in the magnetic field is \( \boxed{2\%} \).