A coil of 100 turns carries a current of 5, m | vedclass

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

Question

(D) The total magnetic flux linkage $(\Phi_T)$ in a coil is given by the product of the number of turns $(N)$ and the magnetic flux through a single t

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The total magnetic flux linkage $(\Phi_T)$ in a coil is given by the product of the number of turns $(N)$ and the magnetic flux through a single turn $(\phi)$: $\Phi_T = N\phi$.Given: $N = 100$, $\phi = 10^{-5} \, Wb$, and current $i = 5 \, mA = 5 	imes 10^{-3} \, A$.The formula for self-inductance $(L)$ is $\Phi_T = Li$.Substituting the values: $N\phi = Li$.$100 	imes 10^{-5} = L 	imes (5 	imes 10^{-3})$.$10^{-3} = L 	imes 5 	imes 10^{-3}$.$L = \frac{10^{-3}}{5 	imes 10^{-3}} = \frac{1}{5} = 0.2 \, H$.Converting to $mH$: $0.2 \, H = 0.2 	imes 1000 \, mH = 200 \, mH$.Since $200 \, mH$ is not among the options, the correct choice is $D$.

$A$ coil of $100$ turns carries a current of $5\, mA$ and creates a magnetic flux of $10^{-5} \, Wb$. The inductance is.....$mH$.

Similar Questions

The current through a coil of self-inductance $L = 2 \ mH$ is given by $I = t^2 e^{-t}$ at time $t$. How long will it take for the induced electromotive force $(emf)$ to become zero (in $s$)?

When the number of turns and the length of the solenoid are doubled keeping the area of cross-section same, the inductance

The inductance of a close-packed coil of $400$ turns is $8 \, mH$. $A$ current of $5 \, mA$ is passed through it. The magnetic flux through each turn of the coil is

$A$ coil of $Cu$ wire (radius $r$,self-inductance $L$) is bent into two concentric turns,each having a radius of $r/2$. What is the new self-inductance?

Find out the $e.m.f.$ produced when the current changes from $0$ to $1 \, A$ in $10 \, s$,given $L = 10 \, \mu H$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module