TS EAMCET · Physics · Electromagnetic Induction
Consider a toroid with rectangular cross section, of inner radius \(a\), outer radius \(b\) and height \(h\), carrying \(n\) number of turns. Then the self-inductance of the toroidal coil when current \(I\) passing through the toroid is
- A \(\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)\)
- B \(\frac{\mu_0 n h}{2 \pi} \ln \left(\frac{b}{a}\right)\)
- C \(\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{a}{b}\right)\)
- D \(\frac{\mu_0 n h}{2 \pi} \ln \left(\frac{a}{b}\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu_0 n^2 h}{2 \pi} \ln \left(\frac{b}{a}\right)\)
Step-by-step Solution
Detailed explanation
Given, a toroid with a rectangular cross-section of inner radius \(a\) and outer radius \(b\). Height of the solenoid \(=h\) Magnetic field inside a rectangular toroid is given by \(B=\frac{\mu_0 n I}{2 \pi r}\) using the infinitesimal cross-sectional area element, \(d x=h d r\)…
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