Question

A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by one-fifth, with the geometric dimensions kept the same. By how much must the current change if the energy stored in the inductor is to remain the same? Select one: a. It must be two times larger. b. It should be one-half as large. c. It should be left the same. d. It must be four times larger. e. No change in the current can compensate for the reduction in the number of turns.

Answer 1

The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

Explanation:

A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:

$U= \frac{1}{2} LI^2$

Here L is the inductance and I is the current.

Here, energy stored (U) is proportional to the number of turns (N) and the current (I).

$L = \frac{\mu_0 N^2 *A}{l}$

mu not - permeability of core material

A -area of cross section

l - length

N - no. of turns in solenoid inductor

Now,given that the proportion always remains same:

$\frac{N_2}{N_1} = \frac{I_1}{I_2}$

In this way the expression

$\frac{1}{5} = \frac{I_1}{I_2}$

$I_2 = I_1 \times 5$

Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.