# If the current flowing in a coil is reduced to half of its original value,

If the current flowing in a coil is reduced to half of its original value, the relation between the new energy (E2) and the original energy (E1) stored in the coil will be

(A) E2 = E1

(B) E2 = 2E1

(C) E2 = $$1\over2$$E1

(D) E2 = $$1\over4$$E1

verified

The Correct option is  D)  E2 = $$1\over4$$E

Explaination

Current is become half of the given value.

$$E={1\over2}Li^2$$

i.e.

$${E_2\over E_1}={({i_2\over i_1})^2}=({1\over2})^2$$         __{current halfs i2=1/2 i1}

$$E_2={1\over4}E_1$$