Question

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.250 A/s, the induced emf in the second coil has a magnitude of 1.65 x 10^-3 V. a) What is the mutual inductance of the pair of coils? b) If the second coil has 30 turns, what is the flux through each turn when the current in the first coil equals 1.25 A? c) If the current in the second coil increases at a rate of 0.3 A/s, what is the magnitude of the induced emf in the first coil?

Answer 1

Explanation:

Given that,

Induced emf $\epsilon= 1.65\times10^{-3}\ V$

Rate of current = 0.250 A/s

Number of turns =30

(a). We need to calculate the mutual inductance of the pair of coils

Using formula of the mutual inductance

$M=\dfrac{\epsilon}{|\dfrac{\Delta i}{\Delta t}|}$

$M=\dfrac{1.65\times10^{-3}}{0.250}$

$M=0.0066=6.6\times10^{-3}\ Hz$

$M=6.6\ mHz$

The mutual inductance of the pair of coils is 6.6 mHz.

(b). We need to calculate the flux through each turn

Using formula of flux

$\phi_{B}=\dfrac{Mi}{N}$

Put the value into the formula

$\phi_{B}=\dfrac{6.6\times10^{-3}\times1.25}{30}$

$\phi_{B}=0.000275 =2.75\times10^{-4}\ Wb$

The flux through each turn is $2.75\times10^{-4}\ Wb$

(c). We need to calculate the magnitude of the induced emf in the first coil

Using formula of induced emf

$\epsilon=M|\dfrac{\Delta i_{2}}{\Delta t}|$

$\epsilon=6.6\times10^{-3}\times0.3$

$\epsilon=0.00198 =1.98\times10^{-3}\ mV$

The magnitude of the induced emf in the first coil is $1.98\times10^{-3}\ mV$

Hence, This is the required solution.

Answer 2

Answer:

Explanation:

We have two which are wound around in the same form

The rate decrease of current in the first coils is

di₁/dt = —0.25A/s

Induced emf in the second coil is

ε₂ = 1.65 × 10^-3 V

A. Mutual inductance?

The induced emf in the second coil in terms of the mutual inductance and the rate of the change in current is given as

ε₂ = M•|di₁/dt|

Where M is mutual inductance

Making M subject of formula

M = ε₂ / di₁/dt

Then, M = 1.65 × 10^-3 / 0.25

M = 6.6 × 10^-3 H

B. If the second coil has a turn of 30turns

N₂ = 30turns

Also, current in the first coil is

i₁ = 1.25A

Flux Φ?

The average flux in the second coil is given as

Φ₂ = M•i₁ / N₂

Φ₂ = 6.6 × 10^-3 × 1.25 / 30

Φ₂ = 2.75 × 10^-4 Wb

C. The current in the second coil increase at rate of

di₂/dt = 0.3 A/s

Induce emf in coil 1 ε₁?

ε₁ = M•|di₂/dt|

ε₁ = 6.6 × 10^-3 × 0.3

ε₁ = 1.98 × 10^-3 V