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:

When a material with magnetic susceptibility \( \chi \) is inserted into a toroidal coil, the magnetic field increases by a factor of \( (1 + \chi) \), and the percentage increase is \( \chi \times 100 \).

Answer 1

The Correct Option is C

Step 1: Understanding the magnetic field inside the toroid.
The magnetic field \( B \) inside a toroid with current \( I \) is given by the formula:
\[ B = \frac{\mu_0 N I}{2 \pi r}, \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns of the coil,
- \( I \) is the current passing through the coil,
- \( r \) is the radius of the toroid.
When the toroid is filled with a material of magnetic susceptibility \( \chi \), the permeability \( \mu \) of the material becomes:
\[ \mu = \mu_0 (1 + \chi). \]

Step 2: Magnetic field with material inside the toroid.
The magnetic field in the toroid with the material inserted becomes:
\[ B = \frac{\mu N I}{2 \pi r} = \frac{\mu_0 (1 + \chi) N I}{2 \pi r}. \]

Step 3: Comparing the magnetic fields.
The magnetic field without the material is:
\[ B_0 = \frac{\mu_0 N I}{2 \pi r}. \]
The ratio of the magnetic field with the material to the magnetic field without the material is:
\[ \frac{B}{B_0} = \frac{\mu_0 (1 + \chi) N I}{2 \pi r} \div \frac{\mu_0 N I}{2 \pi r} = 1 + \chi. \]

Step 4: Percentage increase in the magnetic field.
The percentage increase in the magnetic field is given by:
\[ \text{Percentage increase} = (1 + \chi - 1) \times 100 = \chi \times 100. \]
Final Answer:
Thus, the percentage increase in the magnetic field is:
\[ \boxed{(1 + \chi) \times 100}. \]