A coil of inductance `1 H` and resistance `10Omega` is connected to a resistanceless battery of emf `50 V` at time `t=0`. Calculate the ratio of rthe rate which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at `t=0.1s`.

Explanation

The current growth is given by i=`i_(0)(I-e^(-t//tau))`
`tau= (L)/( R)=(1)/(10) ` sec
`(di)/(dt)= (i_(0))/(tau) e^(-t//tau)` and `i_(0)=( E)/(R )= (50)/(10)= 5 ` amp .
Magnetic field energy in the coil `U_(B)= (1)/(2) Li^(2)`
Rate of increase of magnetic field energy
`(du_(B))/(dt)=Li""(di)/(dt)= (Li_(0)^(2))/(tau) (1-e^(-t//tao))e^(-t//tau) = 250 (1-e^(-1))e^(-1)= 58.12`
Rate of heat produced `i^(2)R = i_(0)^(2)(1-e^(-t//tau))^(2)` R
`= 5^(2)xx10xx(1-e^(-10xx1))^(2) `= 158.0
`:.` Rate of increase of magnetic field energy `(58.12)/(158)=0.37`