Question

square coil of 10cm side and with 100turns is rotated at a uniform speed of 500rpm about an axis at right angle to a uniform speed of 0.5Wb/m?. Calculate the instantancous value of induced e.m.f when the plane of the coil is (i) at the right angle to the plane of the field (i) in the plane of the field (ill) at 45 with the field direction.

Answer 1

The induced e.m.f at the right angle to the plane of the field is zero. The induced e.m.f in the plane of the field is -142.45 and The induced e.m.f at 45 with the field direction is -116.31

To calculate the instantaneous value of the induced electromotive force (e.m.f.) in each of the given orientations of the coil, we can use Faraday's law of electromagnetic induction, which states that the induced e.m.f.
`(\( \varepsilon \))` is equal to the negative rate of change of magnetic flux through the coil.

The formula for the induced e.m.f.
`(\( \varepsilon \))` is given by:

`\[ \varepsilon = -N (d\Phi)/(dt) \]`

Where:

-
`\( \varepsilon \)` is the induced e.m.f.

-
`\( N \)` is the number of turns in the coil.

-
`\( (d\Phi)/(dt) \)` is the rate of change of magnetic flux.

The magnetic flux
`(\( \Phi \))` through a coil is given by the product of the magnetic field
`(\( B \))`, the area
`(\( A \))` of the coil, and the cosine of the angle
`(\( \theta \))` between the normal to the coil and the magnetic field:

`\[ \Phi = B \cdot A \cdot \cos(\theta) \]`

Now, let's calculate the instantaneous values of the induced e.m.f. for the three given orientations:

(i) At right angles to the plane of the field
`(\( \theta = 90^\circ \))`:

`\[ \varepsilon_1 = -N (d(B \cdot A \cdot \cos(90^\circ)))/(dt) \]`

(ii) In the plane of the field
`(\( \theta = 0^\circ \))`:

`\[ \varepsilon_2 = -N (d(B \cdot A \cdot \cos(0^\circ)))/(dt) \]`

(iii) At 45 degrees with the field direction
`(\( \theta = 45^\circ \))`:

`\[ \varepsilon_3 = -N (d(B \cdot A \cdot \cos(45^\circ)))/(dt) \]`

Given values:

- Side length of the square coil
`(\( a \))` = 10 cm = 0.1 m

- Number of turns
`(\( N \))` = 100

- Magnetic field strength
`(\( B \))` = 0.5 Wb/m²

- Angular speed
`(\( \omega \))` =
`\( (500 * 2\pi)/(60) \)` radians/second

The area
`(\( A \))` of the coil is
`\( a^2 \)`.

Now, calculate the instantaneous values for each orientation.

(i) At right angles to the plane of the field
`(\( \theta = 90^\circ \))`:

`\[ \varepsilon_1 = -N (d(B \cdot A \cdot \cos(90^\circ)))/(dt) \]`

Since
`\( \cos(90^\circ) = 0 \)`, the induced e.m.f.
`(\( \varepsilon_1 \))` for this orientation is zero.

(ii) In the plane of the field
`(\( \theta = 0^\circ \))`:

`\[ \varepsilon_2 = -N (d(B \cdot A \cdot \cos(0^\circ)))/(dt) \]`

Since
`\( \cos(0^\circ) = 1 \)`, the induced e.m.f.
`(\( \varepsilon_2 \))` for this orientation is given by:

`\[ \varepsilon_2 = -N (d(B \cdot A))/(dt) \]`

Substitute the values:

`\[ \varepsilon_2 = -100 (d(0.5 \cdot 0.1^2))/(dt) \]`

(iii) At 45 degrees with the field direction
`(\( \theta = 45^\circ \))`:

`\[ \varepsilon_3 = -N (d(B \cdot A \cdot \cos(45^\circ)))/(dt) \]`

Since
`\( \cos(45^\circ)` =
`(√(2))/(2) \)`, the induced e.m.f.
`(\( \varepsilon_3 \))` for this orientation is given by:

`\[ \varepsilon_3 = -100 (d(0.5 \cdot 0.1^2 \cdot (√(2))/(2)))/(dt) \]`

Now, differentiate the expressions for
`\( \varepsilon_2 \)` and
`\( \varepsilon_3 \)` with respect to time and substitute the values to get the final calculations. Note that
`\( \omega \)` is the angular speed.

`\[ \varepsilon_2 = -100 \cdot 2 \pi \omega \cdot 0.5 \cdot 0.1^2 \]`

`\[ \varepsilon_3 = -100 \cdot 2 \pi \omega \cdot 0.5 \cdot 0.1^2 \cdot (√(2))/(2) \]`

Now, substituting the given values for
`\( \omega \)` we get:

`\[ \varepsilon_2 = -100 \cdot 2 \pi \left((500 \pi)/(30)\right) \cdot 0.5 \cdot 0.1^2 \]`

`\[ \varepsilon_3 = -100 \cdot 2 \pi \left((500 \pi)/(30)\right) \cdot 0.5 \cdot 0.1^2 \cdot (√(2))/(2) \]`

`\[ \varepsilon_2 =-142.45 \]`

`\[ \varepsilon_3 = -116.31`