A toroid is a long coil of wire (N turns) wou | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$,where $\phi$ is the total magnetic flux linked with the coil and $I$ is t

Vedclass · Accepted Answer

$\frac{\mu_0 N^2 r^2}{2 R}$

Vedclass · Answer

(D) The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$,where $\phi$ is the total magnetic flux linked with the coil and $I$ is the current flowing through it.For a toroid with $N$ turns,the total flux $\phi$ is given by $\phi = N \cdot A \cdot B$,where $A$ is the cross-sectional area of the toroid and $B$ is the magnetic field inside it.The cross-sectional area $A = \pi r^2$.The magnetic field inside a toroid is $B = \mu_0 n I$,where $n = \frac{N}{2 \pi R}$ is the number of turns per unit length.Substituting these into the flux equation:$\phi = N (\pi r^2) (\mu_0 \frac{N}{2 \pi R} I) = \frac{\mu_0 N^2 r^2 I}{2 R}$.Finally,the self-induction $L = \frac{\phi}{I} = \frac{\mu_0 N^2 r^2}{2 R}$.

$A$ toroid is a long coil of wire ($N$ turns) wound over a circular core. The coefficient of self-induction of the toroid is [The magnetic field in it is uniform and $R >> r$,where $r=$ radius of wire,$R=$ radius of coil] ($\mu_0=$ permeability of free space).

Similar Questions

When current $I$ passes through an inductor having self-inductance of $4 \ H$. If the current is made double,what will be the new self-inductance of the inductor?

$A$ coil has $N$ turns and the current passing through it is $I$ ampere, resulting in a self-inductance of $L$ Henry. If the current is doubled, the new self-inductance will be . . . . . . $H$.

The length of thin wire required to manufacture a solenoid of inductance $L$ and length $\ell$ (if cross-sectional diameter is considered much less than its length) is:

$A$ solenoid is tightly wound with wire of diameter $0.10\,cm$,has a cross-sectional area $\frac{1}{\pi}\,cm^2$,and is $40\,cm$ long. If the current through the solenoid decreases uniformly from $10\,A$ to $0\,A$ in $0.10\,s$,then the $emf$ induced inside the solenoid is.....$mV$.

The self-inductance of an air-core inductor (solenoid) is $0.03 \ mH$. By introducing an iron core into the inductor, the self-inductance increases to $30 \ mH$. The relative permeability of the core used is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module