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

The number of turns in the coil is determined using Faraday's Law of Electromagnetic Induction. This law states that the induced electromotive force (emf), denoted by \(\varepsilon\), in a coil is directly proportional to the rate of change of magnetic flux through it. The governing equation is:

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

Here:

• \(\varepsilon\) represents the induced emf.
• \(N\) is the number of turns in the coil.
• \(\Delta \Phi\) signifies the change in magnetic flux.
• \(\Delta t\) is the duration over which the change occurs.

Magnetic flux (\(\Phi\)) through the coil is calculated as:

\(\Phi = B \cdot A\)

Where:

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

The coil's area is derived from its diameter. Given a diameter \(d = 0.02 \, \text{m}\), the radius is \(r = \frac{d}{2} = 0.01 \, \text{m}\). Consequently, 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 calculated as:

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

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

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

With a time duration \(\Delta t = 2 \, \text{s}\) and an induced emf \(\varepsilon = 22 \, \text{V}\), the equation becomes:

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

Rearranging to solve for \(N\):

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

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

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

Upon calculation:

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

Therefore, the coil possesses 70 turns.