A solenoid has a turn density of 5000 , text{ | vedclass

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

Question

(A) The energy stored in a solenoid is given by $U = \frac{1}{2} LI^2$.The self-inductance $L$ of a solenoid is $L = \frac{\mu_0 \mu_r N^2 A}{\ell}$,

Vedclass · Accepted Answer

$15.7$

Vedclass · Answer

(A) The energy stored in a solenoid is given by $U = \frac{1}{2} LI^2$.The self-inductance $L$ of a solenoid is $L = \frac{\mu_0 \mu_r N^2 A}{\ell}$, where $N$ is the total number of turns and $\ell$ is the length.Energy per unit length $u = \frac{U}{\ell} = \frac{1}{\ell} \left( \frac{1}{2} \frac{\mu_0 \mu_r N^2 A}{\ell} I^2 ight) = \frac{1}{2} \mu_0 \mu_r \left( \frac{N}{\ell} ight)^2 A I^2$.Given: turn density $n = \frac{N}{\ell} = 5000 \, 	ext{turns/m}$, area $A = 10 \, 	ext{cm}^2 = 10 	imes 10^{-4} \, 	ext{m}^2$, current $I = 1 \, 	ext{A}$, relative permeability $\mu_r = 1000$, and $\mu_0 = 4\pi 	imes 10^{-7} \, 	ext{T} \cdot 	ext{m/A}$.Substituting the values:$u = \frac{1}{2} 	imes (4\pi 	imes 10^{-7}) 	imes 1000 	imes (5000)^2 	imes (10 	imes 10^{-4}) 	imes (1)^2$$u = \frac{1}{2} 	imes 4\pi 	imes 10^{-7} 	imes 10^3 	imes 25 	imes 10^6 	imes 10^{-3} 	imes 1$$u = 2\pi 	imes 10^{-7} 	imes 25 	imes 10^9 	imes 10^{-3} = 50\pi \approx 157.08 	imes 0.1 = 15.7 \, 	ext{J/m}$.

$A$ solenoid has a turn density of $5000 \, \text{turns/m}$ and a cross-sectional area of $10 \, \text{cm}^2$. If a current of $1 \, \text{A}$ flows through it and the core material has a relative permeability of $1000$, find the energy per unit length of the solenoid in $\text{J/m}$.

Similar Questions

There are $50$ turns per $cm$ length in a very long solenoid. It carries a current of $2.5 \ A$. The magnetic field at its centre on the axis is . . . . . . $T$.

$A$ current $I$ flows along the length of an infinitely long,straight and thin-walled pipe. Then

An infinitely long hollow conducting cylinder with radius $R$ carries a uniform current along its surface. Choose the correct representation of magnetic field $(B)$ as a function of radial distance $(r)$ from the axis of the cylinder.

The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

$A$ thick current-carrying cable of radius $R$ carries current $I$ uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance $r$ from the axis of the cable is represented by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module