An air-cored solenoid with length 30 cm,area | vedclass

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

Question

(B) The magnetic field $B$ at the center of the solenoid is given by $B = \mu_0 (N/l) i$.The magnetic flux $\Phi$ linked with the solenoid is $\Phi =

Vedclass · Accepted Answer

$6.5$

Vedclass · Answer

(B) The magnetic field $B$ at the center of the solenoid is given by $B = \mu_0 (N/l) i$.The magnetic flux $\Phi$ linked with the solenoid is $\Phi = N B A = N (\mu_0 N i A / l) = (\mu_0 N^2 A / l) i$.The self-inductance $L$ is defined as $L = \Phi / i = \mu_0 N^2 A / l$.Substituting the given values: $N = 500$,$l = 0.3 \ m$,$A = 25 	imes 10^{-4} \ m^2$,$\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$.$L = \frac{4\pi 	imes 10^{-7} 	imes (500)^2 	imes 25 	imes 10^{-4}}{0.3} \approx 2.618 	imes 10^{-3} \ H$.The average back e.m.f. is $e = L (\Delta i / \Delta t)$.Given $\Delta i = 2.5 \ A$ and $\Delta t = 10^{-3} \ s$.$e = (2.618 	imes 10^{-3}) 	imes (2.5 / 10^{-3}) = 2.618 	imes 2.5 \approx 6.545 \ V$.Rounding to the nearest value,$e \approx 6.5 \ V$.

Similar Questions

When the number of turns in a solenoid are doubled without any change in the length of the solenoid,its self-inductance becomes

When the current changes from $+2 \, A$ to $-2 \, A$ in $0.05 \, s$,an $e.m.f.$ of $8 \, V$ is induced in a coil. The coefficient of self-induction of the coil is ... $H$.

When a current $i$ through a solenoid is increasing at a constant rate, then the induced current is

If in a coil,the rate of change of area is $5 \, m^2/ms$ and the current changes from $2 \, A$ to $1 \, A$ in $2 \times 10^{-3} \, s$. If the magnitude of the magnetic field is $1 \, T$,then the self-inductance of the coil is ... $H$.

The inductance of a coil is $60\,\mu H$. $A$ current in this coil increases from $1.0\,A$ to $1.5\,A$ in $0.1\,s$. The magnitude of the induced $e.m.f.$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module