Question

Two coils are at fixed locations. When coil 1 has no current and the current in coil 2 increases at the rate 15.0 A/s, the emf in coil 1 is 25.0 mV. (a) What is their mutual inductance? (b) When coil 2 has no current and coil 1 has a current of 3.60A, what is the flux linkage in coil 2?

Short answer

a)$M=1.67\times {10}^{-3}\mathrm{H}$

b)${\varphi }_{21}=6.0\times {10}^{-3}Wb$

Step-by-step solution

Given

Rate of change of current,

$$\begin{gathered} \frac{d{i}_2}{dt}=15.0A/s \\ {\epsilon }_1=25.0mV=25.0\times {10}^{-3}V \\ {I}_1=3.6A \end{gathered}$$

Understanding the concept

We have the formula for the mutual inductance induced in the coil, we also have enough information given, so we can calculate the mutual inductance. We also have enough information along with the formula for flux linkage, so we can find the required answer.

Formula:

$\varphi =\frac{MI}{N}$

$M=\frac{\epsilon }{\left|di/dt\right|}$

(a) Calculate the mutual inductance

We have,

$M=\frac{\epsilon }{\left|{\displaystyle \frac{di}{dt}}\right|}$

So, mutual inductance in coil 1 will become,

$$\begin{gathered} M=\frac{\epsilon }{\left|{\displaystyle \frac{di}{dt}}\right|} \\ M=\frac{25.0\times {10}^{-3}}{15} \\ M=1.67+{10}^{-3}H \end{gathered}$$

(b) Calculate the flux linkage in coil 2 when coil 2 has no current and coil 1 has a current of 3.60 A

We have,

$\varphi =\frac{MI}{N}$

Here, N=1

$$\begin{gathered} {\varphi }_{21}=\frac{M{I}_1}{N} \\ {\varphi }_{21}=\frac{1.67\times {10}^{-3}\times 3.6}{1} \\ {\varphi }_{21}=6.0\times {10}^{-3}Wb \end{gathered}$$