Question

A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is :

• A. 7
• B. 70
• C. 35
• D. 140

Answer 1

The Correct Option is B

To find the number of turns in the coil given the information about the change in the magnetic field, we can use Faraday's Law of Electromagnetic Induction, which states that the induced emf (\(\varepsilon\)) in a coil is proportional to the rate of change of magnetic flux through the coil. The formula is:

\(\varepsilon = -N \frac{\Delta \Phi}{\Delta t}\) 

where:

• \(\varepsilon\) is the induced emf,
• \(N\) is the number of turns in the coil,
• \(\Delta \Phi\) is the change in magnetic flux,
• \(\Delta t\) is the time duration of the change.

The magnetic flux \(\Phi\) through the coil is given by:

\(\Phi = B \cdot A\)

where:

• \(B\) is the magnetic field,
• \(A\) is the area of the coil.

The area \(A\) of the coil can be calculated from its diameter. Given the diameter \(d = 0.02 \, \text{m}\), the radius \(r = \frac{d}{2} = 0.01 \, \text{m}\). Thus, the area is:

\(A = \pi r^2 = \pi (0.01)^2 = 0.0001\pi \, \text{m}^2\)

The change in magnetic field \(\Delta B\) is:

\(\Delta B = B_{\text{final}} - B_{\text{initial}} = 3000 \, \text{T} - 5000 \, \text{T} = -2000 \, \text{T}\)

The change in magnetic flux \(\Delta \Phi\) is:

\(\Delta \Phi = A \cdot \Delta B = 0.0001\pi \cdot (-2000) = -0.2\pi \, \text{Wb}\)

The time duration \(\Delta t = 2 \, \text{s}\). Since the induced emf \(\varepsilon = 22 \, \text{V}\), we can write:

\(22 = -N \cdot \frac{-0.2\pi}{2}\)

Solving for \(N\):

\(22 = N \cdot \frac{0.2\pi}{2}\)

\(22 = N \cdot 0.1\pi\)

\(N = \frac{22}{0.1\pi}\)

Calculating the value:

\(N \approx \frac{22}{0.314} \approx 70\)

Thus, the number of turns in the coil is 70.

Answer 2

The induced emf in the coil is given by:

\[ \mathcal{E} = N \left( \frac{\Delta \phi}{t} \right) \]

Where:

\[ \Delta \phi = (\Delta B)A \]

Given:

\[ B_i = 5000 \, \text{T}, \quad B_f = 3000 \, \text{T} \] \[ d = 0.02 \, \text{m} \implies r = 0.01 \, \text{m} \]

Calculating the change in magnetic flux:

\[ \Delta \phi = (\Delta B)A = (2000)(\pi)(0.01)^2 = 0.2\pi \]

Substituting values into the emf equation:

\[ \mathcal{E} = N \left( \frac{0.2\pi}{2} \right) \]

Given \( \mathcal{E} = 22 \, \text{V} \), we get:

\[ 22 = N \left( \frac{0.2\pi}{2} \right) \]

Solving for \( N \):

\[ N = 70 \]