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?

The current must be 5 times larger

Explanation:

The equation for Inductance, is given by;

L = (μ_o•N²•A)/l

Where;

μ_o is the permeability of the core material

N is the number of turns

A is the cross sectional Area

l is the length of the coil in meters

L is the inductance

However, we also need to find an expression for the Energy Stored, and the equation is given by;

U = (1/2)LI²

Where;

L is the inductance

I is the current.

We also know that U is proportional to the number of turns and the current.

Thus, we can conclude that;

N_2/N_1 = I_1/I_2

From the question, we are told that the number of turns is reduces by 1/5. Thus,

1/5 = I_1/I_2

So, I_2 = 5I_1

So the current must be 5 times larger

v=ir = 3x15 =45 v

grav potential energy