Question

The space within the current carrying toroid is filled with aluminium of susceptibility \( \chi \). The percentage increase in the magnetic field \( B \) will be

• A. \( \frac{3}{2} \times 100 \)
• B. \( 2\chi \times 100 \)
• C. \( (1 + \chi) \times 100 \)
• D. \( \chi \times 100 \)

Hint:

The field inside a toroid with a core is multiplied by the relative permeability, which is one plus the susceptibility.

Answer 1

The Correct Option is D

Step 1: Magnetisation
The core gets magnetised with $M = \chi H$, and $B = \mu_0(H + M) = \mu_0 H(1 + \chi)$.

Step 2: Compare with the Empty Toroid
The empty toroid has $B_0 = \mu_0 H$, so $\dfrac{B}{B_0} = 1 + \chi$.

Step 3: Compare the options
The fractional increase is $\chi$, so the percentage increase is $\chi \times 100$, the last option.

Final Answer:
The field increases by chi times 100 percent. \[ \boxed{\text{(D) }\chi \times 100} \]