Two coils $$ (X) $$ and $$ (Y) $$, the number of turns of coil $$ (X) $$ is 1000 and number of turns of coil (Y) 500 , when the current changes in coil $$ x $$ from 0 to 5 A during time 0.2 sec a magnetic flux 8 mWb cuts the coil $$ Y $$, so the mutual induction coefficient equals......... $$ \begin{array}{llll} ext { a) } 0.8 \mathrm{H} & ext { b) } 8 \mathrm{H} & ext { c) } 0.16 \mathrm{H} & ext { d) } 16 \mathrm{H}\end{array} $$

Question

Two coils $$ (X) $$ and $$ (Y) $$, the number of turns of coil $$ (X) $$ is 1000 and number of turns of coil (Y) 500 , when the current changes in coil $$ x $$ from 0 to 5 A during time 0.2 sec a magnetic flux 8 mWb cuts the coil $$ Y $$, so the mutual induction coefficient equals......... $$ \begin{array}{llll}	ext { a) } 0.8 \mathrm{H} & 	ext { b) } 8 \mathrm{H} & 	ext { c) } 0.16 \mathrm{H} & 	ext { d) } 16 \mathrm{H}\end{array} $$

UpStudy AI · Accepted Answer

To find the mutual induction coefficient $$ M $$ between the two coils $$ (X) $$ and $$ (Y) $$, we can use the formula:

$$
M = \frac{\Phi}{\Delta I / \Delta t}
$$

where:
- $$ \Phi $$ is the magnetic flux through coil $$ (Y) $$ in Weber (Wb),
- $$ \Delta I $$ is the change in current in coil $$ (X) $$ in Amperes (A),
- $$ \Delta t $$ is the time interval in seconds (s).

### Given Data:
- Number of turns in coil $$ (X) $$: $$ N_X = 1000 $$
- Number of turns in coil $$ (Y) $$: $$ N_Y = 500 $$
- Change in current $$ \Delta I = 5 \, \text{A} - 0 \, \text{A} = 5 \, \text{A} $$
- Time interval $$ \Delta t = 0.2 \, \text{s} $$
- Magnetic flux $$ \Phi = 8 \, \text{mWb} = 8 \times 10^{-3} \, \text{Wb} $$

### Step 1: Calculate $$ \Delta I / \Delta t $$

$$
\frac{\Delta I}{\Delta t} = \frac{5 \, \text{A}}{0.2 \, \text{s}} = 25 \, \text{A/s}
$$

### Step 2: Calculate the mutual induction coefficient $$ M $$

Now we can substitute the values into the formula for $$ M $$:

$$
M = \frac{8 \times 10^{-3} \, \text{Wb}}{25 \, \text{A/s}}
$$

### Step 3: Perform the calculation

Now, let's calculate $$ M $$:

$$
M = \frac{8 \times 10^{-3}}{25}
$$

Calculating this gives:

$$
M = 0.00032 \, \text{H} = 0.32 \, \text{H}
$$

However, we need to consider the number of turns in coil $$ (Y) $$ since the mutual induction coefficient is also affected by the number of turns in the secondary coil. The formula can be adjusted to:

$$
M = \frac{N_Y \cdot \Phi}{\Delta I / \Delta t}
$$

Substituting $$ N_Y = 500 $$:

$$
M = \frac{500 \cdot 8 \times 10^{-3}}{25}
$$

Calculating this gives:

$$
M = \frac{4}{25} = 0.16 \, \text{H}
$$

### Final Answer

Thus, the mutual induction coefficient $$ M $$ equals $$ 0.16 \, \text{H} $$.

The correct option is **c) 0.16 H**.
The mutual induction coefficient $$ M $$ is $$ 0.16 \, \text{H} $$.

Two coils and , the number of turns of coil is 1000 and number of turns of coil (Y)
500 , when the current changes in coil from 0 to 5 A during time 0.2 sec a magnetic flux 8
mWb cuts the coil , so the mutual induction coefficient equals…

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Two coils and , the number of turns of coil is 1000 and number of turns of coil (Y) 500 , when the current changes in coil from 0 to 5 A during time 0.2 sec a magnetic flux 8 mWb cuts the coil , so the mutual induction coefficient equals…