Question

A toroid has a non-ferromagnetic core of inner radius \(24\;cm\) and outer radius \(25\;cm\), around which \(4900\) turns of a wire are wound. If the current in the wire is \(12\;A\), the magnetic field inside the core of the toroid is

• A. \(56\;mT\)
• B. \(54\;mT\)
• C. \(42\;mT\)
• D. \(48\;mT\)

Hint:

For a toroid, \[ B=\frac{\mu_0NI}{2\pi r} \] where \(r\) is the mean radius of the toroid.

Answer 1

The Correct Option is D

Step 1: Recall the toroid field formula.
For a toroid with a non-ferromagnetic core, the field along the core is \[ B = \frac{\mu_0 N I}{2\pi r} \] where $r$ is the mean radius.
Step 2: Compute the mean radius.
With inner radius $24\ \text{cm}$ and outer radius $25\ \text{cm}$, \[ r = \frac{24 + 25}{2} = 24.5\ \text{cm} = 0.245\ \text{m} \]
Step 3: List the remaining data.
Here $N = 4900$ turns, $I = 12\ \text{A}$, and $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$.
Step 4: Substitute into the formula.
\[ B = \frac{(4\pi \times 10^{-7})(4900)(12)}{2\pi (0.245)} \]
Step 5: Simplify by cancelling $\pi$.
The $\pi$ cancels and the factor $\dfrac{4}{2} = 2$ remains, giving \[ B = \frac{2 \times 10^{-7} \times 4900 \times 12}{0.245} \]
Step 6: Evaluate the number.
The numerator is $2 \times 10^{-7} \times 58800 = 1.176 \times 10^{-2}$, and dividing by $0.245$ gives \[ B \approx 0.048\ \text{T} = 48\ \text{mT} \] so \[ \boxed{48\ \text{mT}} \]