A toroid is a long coil of wire wound over a | vedclass

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

Question

(C) The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$,where $\phi$ is the magnetic flux and $I$ is the current.For a toroid,the

Vedclass · Accepted Answer

$\frac{\mu_{0} N^{2} r^{2}}{2 R}$

Vedclass · Answer

(C) The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$,where $\phi$ is the magnetic flux and $I$ is the current.For a toroid,the magnetic field $B$ inside the core is given by $B = \mu_{0} n I$,where $n$ is the number of turns per unit length.Given $n = \frac{N}{2 \pi R}$,we have $B = \mu_{0} \left( \frac{N}{2 \pi R} \right) I$.The magnetic flux $\phi$ through each turn of the coil is $\phi = B \cdot A$,where $A = \pi r^{2}$ is the cross-sectional area of the coil.Total flux linked with $N$ turns is $\Phi = N \phi = N (B \cdot A) = N \left( \mu_{0} \frac{N}{2 \pi R} I \right) (\pi r^{2})$.Simplifying this,we get $\Phi = \frac{\mu_{0} N^{2} r^{2} I}{2 R}$.Therefore,the self-inductance $L = \frac{\Phi}{I} = \frac{\mu_{0} N^{2} r^{2}}{2 R}$.

Similar Questions

$A$ long,straight wire of radius $a$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2a$ respectively from the axis of the wire is

$A$ current-carrying solenoid is placed vertically and a particle of mass $m$ with charge $Q$ is released from rest. The particle moves along the axis of the solenoid. If $g$ is the acceleration due to gravity, then the acceleration $(a)$ of the charged particle will satisfy:

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per cm are tripled and the current is made $\left(\frac{1}{4}\right)^{th}$,then the new value of the magnetic field will be:

$A$ long cylindrical wire of radius $R$ carries a uniform current $I$ flowing through it. The variation of magnetic field $B$ with distance $r$ from the axis of the wire is shown by:

$A$ solenoid of length $0.5\; m$ has a radius of $1\; cm$ and is made up of $500$ turns. It carries a current of $5\; A$. What is the magnitude of the magnetic field inside the solenoid?

Vedclass Test Series

Exam Paper Generator

Online Exam Module