A coil having N turns is wound tightly in the | vedclass

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

Question

(C) Consider an element of thickness $dr$ at a distance $r$ from the centre. The number of turns in this element is given by $dN = \left(\frac{N}{b -

Vedclass · Accepted Answer

$\frac{\mu_0 Ni}{2(b - a)} \ln\left(\frac{b}{a}\right)$

Vedclass · Answer

(C) Consider an element of thickness $dr$ at a distance $r$ from the centre. The number of turns in this element is given by $dN = \left(\frac{N}{b - a}\right) dr$.The magnetic field due to this element at the centre of the coil is given by $dB = \frac{\mu_0 (dN) i}{2r} = \frac{\mu_0 i}{2} \left(\frac{N}{b - a}\right) \frac{dr}{r}$.To find the total magnetic field $B$,we integrate $dB$ from $r = a$ to $r = b$:$B = \int_{a}^{b} dB = \frac{\mu_0 Ni}{2(b - a)} \int_{a}^{b} \frac{dr}{r}$.$B = \frac{\mu_0 Ni}{2(b - a)} [\ln r]_{a}^{b} = \frac{\mu_0 Ni}{2(b - a)} \ln\left(\frac{b}{a}\right)$.

$A$ coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $i$ passes through the coil,the magnetic field at the centre is

Similar Questions

Which of the following is a vector quantity?

An infinitely long straight conductor is bent into the shape as shown below. It carries a current of $I$ ampere and the radius of the circular loop is $R$ metre. Then, the magnitude of magnetic induction at the centre of the circular loop is

In terms of basic units of mass $(M)$,length $(L)$,time $(T)$ and charge $(Q)$,the dimensions of magnetic permeability of vacuum $(\mu_0)$ would be

Vedclass Test Series

Exam Paper Generator

Online Exam Module