A current 2 A flows through a solenoid of se | vedclass

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

Question

(B) The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$. Given: Self-inductance $L = 3 \ mH = 3 	imes 10^{-3} \ H$.

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$. Given: Self-inductance $L = 3 \ mH = 3 	imes 10^{-3} \ H$. Current $I = 2 \ A$. Substituting the values into the formula: $U = \frac{1}{2} 	imes (3 	imes 10^{-3} \ H) 	imes (2 \ A)^2$ $U = \frac{1}{2} 	imes 3 	imes 10^{-3} 	imes 4$ $U = 6 	imes 10^{-3} \ J$ Since $1 \ mJ = 10^{-3} \ J$,the energy stored is $6 \ mJ$.

$A$ current $2 \ A$ flows through a solenoid of self-inductance $3 \ mH$. The energy stored inside the solenoid is $.... \ mJ$.

Similar Questions

The magnetic energy stored in an inductor of inductance $5 \mu H$ carrying a current of $2 \ A$ is

Average energy stored in a pure inductance $L$ when a current $i$ flows through it,is

If the current of $I$ gives rise to a magnetic flux $\phi$ through a coil having $N$ turns,then the magnetic energy stored in the medium surrounding the coil is

Energy in a current carrying coil is stored in the form of

The energy stored in a $50 mH$ inductor carrying a current of $4 A$ is (in $J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module