The magnetic flux near the axis and inside th | vedclass

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

Question

(C) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.Magnetic flux $\phi$ through the cross-sectional area $A$ is $\phi

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.Magnetic flux $\phi$ through the cross-sectional area $A$ is $\phi = BA = \frac{\mu_0 NIA}{L}$.The magnetic moment $M$ of the solenoid is defined as $M = NIA$.Substituting $M$ into the flux equation: $\phi = \frac{\mu_0 M}{L}$.Rearranging to solve for $M$: $M = \frac{\phi L}{\mu_0}$.Given $\phi = 1.57 	imes 10^{-6} 	ext{ Wb}$, $L = 0.8 	ext{ m}$, and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T m/A}$.$M = \frac{1.57 	imes 10^{-6} 	imes 0.8}{4 	imes 3.14 	imes 10^{-7}}$.$M = \frac{1.57 	imes 0.8 	imes 10^{-6}}{12.56 	imes 10^{-7}} = \frac{1.256 	imes 10^{-6}}{1.256 	imes 10^{-6}} = 1 	ext{ Am}^2$.

Similar Questions

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

$A$ toroid is:

$A$ current of $1/4\pi \ A$ is flowing through a toroid. It has $1000$ turns per meter. The value of the magnetic field (in $Wb/m^2$) along its axis is:

The free space inside a current-carrying toroid is filled with a material of magnetic susceptibility $2 \times 10^{-2}$. The percentage increase in the value of the magnetic field inside the toroid will be $.....\%$.

$A$ long straight wire of circular cross-section (radius $a$) is carrying steady current $I$. The current $I$ is uniformly distributed across this cross-section. The magnetic field is

Vedclass Test Series

Exam Paper Generator

Online Exam Module