Question

A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by one-half, 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? (a) It must be four times larger. (b) It must be two times larger. (c) It should be left the same. (d) It should be one-half as large. (e) No change in the current can compensate for the reduction in the number of turns.

Answer 1

Option B is correct.

Step-by-step explanation:

We define our equation for Inductance, that is,

`L = (\mu_0 N^2 *A)/(l)`

Where
`\mu_0` is the permeability of the core material

A= is the cross-sectional Area

N is the number of turns

l= is the lenght of the coil in meters

But we need also a expression for the Energy Stored, that equation is given by,

`U = (1)/(2) LI^2`

Where,

L is the inductance and I is the current.

We know that U always is going to be proportional to the number of turns and the current.

Therefore we can conclude that the proportion

`(N_2)/(N_1) = (I_1)/(I_2)`

Always remains same.

In this way the expresion

`(1)/(2)  = (I_1)/(I_2) \Rightarrow I_2 = 2*I_1`

Then it mus be two times larger