A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0220 m. A coil of 100 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system

The inductance is 3.4 X 10⁻³H

Explanation:

Give-

N/L = 1800turns/m

radius, r = 0.022m

N = 100 turns

We know,

Ф = BA

and

B = (N*i*u)/l

Substituting both the equations, we get

Ф = L*i/N

N / (N/l) = l

where,

l = length

L = mutual induction

substitute (N*i*u)/l for B in Ф = BA :

Ф = (N*i*μ*A)/l

set the two Ф equal to each other:

(L*i)/N = (N*i*μ*A)/l

Solve for L:

L = (N^2 X μ X A)/l

Substitute N / (N/l) for l

Therefore, the inductance is 3.4 X 10⁻³H

answer: strength i guess

explanation:

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge