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: Use magnetic field formula for a toroid.
For a toroid with non-ferromagnetic core, \[ B=\frac{\mu_0NI}{2\pi r} \] where \(r\) is the mean radius of the toroid.

Step 2: Find the mean radius.
Inner radius is \[ r_1=24\;cm \] Outer radius is \[ r_2=25\;cm \] Mean radius is \[ r=\frac{r_1+r_2}{2} \] \[ r=\frac{24+25}{2} \] \[ r=24.5\;cm \] \[ r=0.245\;m \]

Step 3: Substitute the given values.
Given, \[ N=4900 \] \[ I=12\;A \] and \[ \mu_0=4\pi\times10^{-7}\;T\,m\,A^{-1} \] So, \[ B=\frac{(4\pi\times10^{-7})(4900)(12)}{2\pi(0.245)} \]

Step 4: Simplify.
\[ B=\frac{2\times10^{-7}\times4900\times12}{0.245} \] \[ B=\frac{117600\times10^{-7}}{0.245} \] \[ B=0.048\;T \] \[ B=48\times10^{-3}\;T \] \[ B=48\;mT \]

Step 5: Final conclusion.
Hence, the magnetic field inside the core of the toroid is \[ \boxed{48\;mT} \]