A toroid has a number of turns per unit lengt | vedclass

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

Question

(A) toroid is essentially a solenoid bent into a circular shape.For a toroid with $n$ turns per unit length and carrying a current $i$,the magnetic fi

Vedclass · Accepted Answer

$\mu_0 ni$

Vedclass · Answer

(A) toroid is essentially a solenoid bent into a circular shape.For a toroid with $n$ turns per unit length and carrying a current $i$,the magnetic field $B$ inside the core is derived using Ampere's Circuital Law.According to Ampere's Law,$\oint B \cdot dl = \mu_0 I_{	ext{enclosed}}$.For a circular path of radius $r$ inside the toroid,the length is $2\pi r$ and the total number of turns is $N = n(2\pi r)$.Thus,$B(2\pi r) = \mu_0 (n \cdot 2\pi r) i$.Simplifying this,we get $B = \mu_0 ni$.

$A$ toroid has a number of turns per unit length $n$ and current $i$. What is the magnetic field inside the toroid?

Similar Questions

What is the magnetic field outside an ideal solenoid?

The magnetic field intensity $(H)$ at the centre of a long solenoid having '$n$' turns per unit length and carrying a current $I$, when no material is kept in it is ($\mu_0$ = permeability of free space)

Give the formula for the magnetic field due to a toroid.

The magnetic field intensity inside a current-carrying solenoid is $H = 2.4 \times 10^3 \ A/m$. If the length and the number of turns of the solenoid are $15 \ cm$ and $60$ turns respectively,the current flowing in the solenoid is: (in $A$)

Which of the following statements is incorrect for a current-carrying solenoid?

Vedclass Test Series

Exam Paper Generator

Online Exam Module