Question

A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of earth in that region is 0.5 G and the angle of dip is \( 30^\circ \). The emf induced across the blades is \( N \pi \times 10^{-5} \, \text{V} \). The value of \( N \) is \( \_\_\_\_\_ \).

Answer 1

Correct Answer: 32

Step 1. Calculate the Effective Vertical Component of the Magnetic Field:

Given:

\( B = 0.5 \, \text{G} = 0.5 \times 10^{-4} \, \text{T} \)

The vertical component of the magnetic field \( B_v \), considering the angle of dip \( \delta = 30^\circ \), is:

\( B_v = B \sin \delta = 0.5 \times 10^{-4} \times \sin 30^\circ = 0.5 \times 10^{-4} \times \frac{1}{2} = \frac{1}{4} \times 10^{-4} \, \text{T} \)

Step 2. Convert Angular Velocity from rpm to rad/s:

Angular velocity \( \omega \) in rad/s is given by:

\( \omega = 2 \pi \times f = 2 \pi \times \frac{1200}{60} = 2 \pi \times 20 = 40 \pi \, \text{rad/s} \)

Step 3. Determine the Radius of Rotation:

The length of each blade is \( \ell = 80 \, \text{cm} = 0.8 \, \text{m} \). Therefore, the effective radius \( r \) of rotation is:

\( r = 0.8 \, \text{m} \)

Step 4. Calculate the Induced emf:

The emf \( \varepsilon \) induced across the tips of the blades (assuming the emf induced across two opposite ends) is given by:

\( \varepsilon = \frac{1}{2} B_v \omega r^2 \)

Substituting the values:

\( \varepsilon = \frac{1}{2} \times \frac{1}{4} \times 10^{-4} \times 40 \pi \times (0.8)^2 \)

Simplifying further:

\( \varepsilon = \frac{1}{2} \times \frac{1}{4} \times 10^{-4} \times 40 \pi \times 0.64 = 32 \pi \times 10^{-5} \, \text{V} \)

Step 5. Conclude the Value of \( N \):

Comparing with \( N \pi \times 10^{-5} \, \text{V} \), we find:

\( N = 32 \)

Answer 2

The problem asks for the value of \(N\), where the electromotive force (emf) induced across the blades of a rotating ceiling fan is given in the form \(N \pi \times 10^{-5} \, \text{V}\).

Concept Used:

The solution is based on the concept of motional emf. When a conductor moves through a magnetic field, an emf is induced across its ends. For a conducting rod of length \(L\) rotating with a constant angular velocity \(\omega\) in a uniform magnetic field \(B\) that is perpendicular to the plane of rotation, the induced emf (\(\mathcal{E}\)) between the center of rotation and the tip of the rod is given by:

\[ \mathcal{E} = \frac{1}{2} B L^2 \omega \]

In this problem, the fan blades are rotating in a horizontal plane. Therefore, the component of the Earth's magnetic field that is perpendicular to the plane of rotation is the vertical component, \(B_V\). The vertical component of the Earth's magnetic field is related to the total magnetic field \(B_E\) and the angle of dip \(\delta\) by:

\[ B_V = B_E \sin(\delta) \]

Step-by-Step Solution:

Step 1: List the given parameters and convert them to SI units.

• Length of each blade, \(L = 80 \, \text{cm} = 0.8 \, \text{m}\).
• Angular velocity, \(\omega = 1200 \, \text{rpm}\) (revolutions per minute). We need to convert this to radians per second. \[ \omega = 1200 \, \frac{\text{rev}}{\text{min}} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 40\pi \, \text{rad/s} \]
• Earth's magnetic field, \(B_E = 0.5 \, \text{G}\) (Gauss). We convert this to Tesla (T). \[ B_E = 0.5 \times 10^{-4} \, \text{T} \]
• Angle of dip, \(\delta = 30^\circ\).

Step 2: Calculate the vertical component of the Earth's magnetic field (\(B_V\)).

This is the component of the magnetic field perpendicular to the plane of rotation of the blades.

\[ B_V = B_E \sin(\delta) = (0.5 \times 10^{-4} \, \text{T}) \times \sin(30^\circ) \] \[ B_V = (0.5 \times 10^{-4}) \times (0.5) = 0.25 \times 10^{-4} \, \text{T} \]

Step 3: Calculate the emf induced across a single blade.

Using the formula for motional emf in a rotating rod:

\[ \mathcal{E} = \frac{1}{2} B_V L^2 \omega \]

Substitute the values:

\[ \mathcal{E} = \frac{1}{2} (0.25 \times 10^{-4}) (0.8)^2 (40\pi) \] \[ \mathcal{E} = \frac{1}{2} (0.25 \times 10^{-4}) (0.64) (40\pi) \]

Final Computation & Result:

Step 4: Simplify the expression for the induced emf and find the value of \(N\).

\[ \mathcal{E} = (0.5 \times 0.25 \times 0.64 \times 40) \pi \times 10^{-4} \] \[ \mathcal{E} = (20 \times 0.25 \times 0.64) \pi \times 10^{-4} \] \[ \mathcal{E} = (5 \times 0.64) \pi \times 10^{-4} \] \[ \mathcal{E} = 3.2 \pi \times 10^{-4} \, \text{V} \]

The problem states that the induced emf is \( N \pi \times 10^{-5} \, \text{V} \). We need to match our result to this format.

\[ 3.2 \pi \times 10^{-4} = 32 \pi \times 10^{-5} \, \text{V} \]

Comparing this with the given expression:

\[ N \pi \times 10^{-5} = 32 \pi \times 10^{-5} \]

This implies that \( N = 32 \).

The value of \(N\) is 32.