Question:
A current of \(\frac{1}{3\pi}\,A\) passes through an ideal toroid of 900 turns per metre. If the relative permeability of the material of the core of the toroid is 400, then the magnetic field inside the core of the toroid is:
A current of \(\frac{1}{3\pi}\,A\) passes through an ideal toroid of 900 turns per metre. If the relative permeability of the material of the core of the toroid is 400, then the magnetic field inside the core of the toroid is:
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For an ideal toroid:
\[
B=\mu_0\mu_r nI.
\]
It is analogous to the magnetic field inside a solenoid.
Updated On: Jun 18, 2026
- \(48mT\)
- \(24mT\)
- \(72mT\)
- \(96mT\)
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The Correct Option is A
Solution and Explanation
Concept:
Magnetic field inside a toroid:
\[
B=\mu_0\mu_r nI.
\]
where
\[
n=\text{turns per metre}.
\]
Step 1: Substitute values.
\[ \mu_0=4\pi\times10^{-7}. \] \[ \mu_r=400. \] \[ n=900. \] \[ I=\frac1{3\pi}. \] \[ B = (4\pi\times10^{-7}) (400) (900) \left(\frac1{3\pi}\right). \]
Step 2: Simplify.
\[ B = 4\times400\times300\times10^{-7}. \] \[ = 4.8\times10^{-2}T. \] \[ = 48\times10^{-3}T. \] \[ =48mT. \] Therefore, \[ \boxed{48mT}. \]
Step 1: Substitute values.
\[ \mu_0=4\pi\times10^{-7}. \] \[ \mu_r=400. \] \[ n=900. \] \[ I=\frac1{3\pi}. \] \[ B = (4\pi\times10^{-7}) (400) (900) \left(\frac1{3\pi}\right). \]
Step 2: Simplify.
\[ B = 4\times400\times300\times10^{-7}. \] \[ = 4.8\times10^{-2}T. \] \[ = 48\times10^{-3}T. \] \[ =48mT. \] Therefore, \[ \boxed{48mT}. \]
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