Chapter 30: Q73P (page 900)

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

Expert verified

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

b) $ϕ_{21} = 6.0 \times 10^{- 3} W b$

Step by step solution

Given

Rate of change of current,

$\frac{d i_{2}}{d t} = 15.0 A / s ε_{1} = 25.0 m V = 25.0 \times 10^{- 3} V I_{1} = 3.6 A$

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:

$ϕ = \frac{M I}{N}$

$M = \frac{ε}{|d i / d t|}$

(a) Calculate the mutual inductance

We have,

$M = \frac{ε}{|\frac{d i}{d t}|}$

So, mutual inductance in coil 1 will become,

$M = \frac{ε}{|\frac{d i}{d t}|} M = \frac{25.0 \times 10^{- 3}}{15} M = 1.67 + 10^{- 3} H$

(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,

$ϕ = \frac{M I}{N}$

Here, N=1

$ϕ_{21} = \frac{M I_{1}}{N} ϕ_{21} = \frac{1.67 \times 10^{- 3} \times 3.6}{1} ϕ_{21} = 6.0 \times 10^{- 3} W b$

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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

Step by step solution

Given

Understanding the concept

(a) Calculate the mutual inductance

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

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