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

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

because if the tool that you are using is not balanced it there would be false mass.