Question

A magnetic field of $2\times 10^{-2}$T acts at right angles to a coil of area 100 $cm^2$, with 50 turns. The average e.m.f. induced in the coil is 0.1 V, when it is removed from the field in t sec. The value of t is

• A. 10 s
• B. 0.1 s
• C. 0.01 s
• D. 1 s

Answer 1

The Correct Option is B

To solve this problem, we need to determine the time \( t \) taken to remove the coil from the magnetic field using the relationship between induced e.m.f., change in magnetic flux, and time.

Step 1: Understanding the Given Values

• Magnetic field strength, \( B = 2 \times 10^{-2} \, \text{T} \).
• Area of the coil, \( A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 10^{-2} \, \text{m}^2 \) (converting square centimeters to square meters).
• Number of turns in the coil, \( N = 50 \).
• Average induced e.m.f., \( \text{e.m.f.} = 0.1 \, \text{V} \).

Step 2: Formula for Induced E.M.F

The induced e.m.f. (\( \varepsilon \)) in a coil when it is removed from a magnetic field can be expressed using Faraday’s law of electromagnetic induction:

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

where \(\Delta \Phi\) is the change in magnetic flux and \(\Delta t\) is the time interval.

Since the coil is completely removed from the field, the final magnetic flux \(\Phi_f = 0\) and the initial flux \(\Phi_i = B \times A\).

Thus, \(\Delta \Phi = \Phi_f - \Phi_i = 0 - B \times A\).

Step 3: Calculating the Change in Magnetic Flux

Substitute \( B \) and \( A \) into the expression for \(\Delta \Phi\):

\(\Delta \Phi = - (2 \times 10^{-2} \, \text{T}) \times (10^{-2} \, \text{m}^2)\)

\(\Delta \Phi = - 2 \times 10^{-4} \, \text{Wb}\)

Step 4: Solving for \( t \)

Substitute \(\Delta \Phi\), \(\text{e.m.f.}\), and \(N\) into Faraday’s law equation:

\(0.1 = 50 \times \frac{-2 \times 10^{-4}}{t}\)

Solve for \( t \):

\(0.1 = -\frac{50 \times 2 \times 10^{-4}}{t}\)

\(0.1 = -\frac{10^{-2}}{t}\)

\(t = -\frac{10^{-2}}{0.1}\)

\(t = 0.1 \, \text{s}\)

Conclusion:

The time taken to remove the coil from the magnetic field is \( 0.1 \, \text{s} \). Thus, the correct answer is 0.1 s.