Two coils of wire are placed close together. Initially, a current of 3.03 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of 3.03 x 10-2 s. During this time, the average emf induced in the other coil is 4.13 V. What is the mutual inductance of the two-coil system

4.13 ×10^-2 H

Explanation:

mutual inductance of the two-coil system can be calculated using below expression

mutual inductance = (- E Δt)/ ΔIp

Where E= average emf induced

Δt= time

ΔIp= change in current

ΔIp=( 0s - 3.03 A)

Δt=3.03 x 10-2 s.

If we substitute the values into above expresion, we have,

E= 4.13 V.

mutual inductance = (- E Δt)/ ΔIp

= - (4.13 V × 3.03 x 10^-2 s.)/ ( 0s - 3.03 A)

= 4.13 ×10^-2 H

Hence, the mutual inductance of the two-coil system is 4.13 ×10^-2 H

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .