A coil of area \(3 \text{ m}^{2}\) is at right angles to a magnetic field of \(0.05 \text{ Wb/m}^{2}\). If the field is decreased to \(20%\) of its value in \(10 \text{ s}\) , the induced e.m.f. is.
Show Hint
E.m.f is the rate of change of flux; flux is $B \times A$ when perpendicular.
Step 1: Understanding the Question:
The magnetic flux through a coil is changing due to a decrease in magnetic field strength. According to Faraday's law, this induces an e.m.f. in the coil. Step 2: Key Formula or Approach:
Induced e.m.f. \(e = -\frac{\Delta \Phi}{\Delta t} = -A \cdot \frac{\Delta B}{\Delta t}\) (since the coil is at right angles, \(\Phi = BA\)).
Magnitude \(|e| = A \cdot \frac{|B_{2} - B_{1}|}{\Delta t}\). Step 3: Detailed Explanation:
Given:
Area \(A = 3 \text{ m}^{2}\)
Initial field \(B_{1} = 0.05 \text{ Wb/m}^{2}\)
Final field \(B_{2} = 20%\) of \(B_{1} = 0.20 \times 0.05 = 0.01 \text{ Wb/m}^{2}\)
Time interval \(\Delta t = 10 \text{ s}\)
Induced e.m.f. magnitude:
\[ |e| = 3 \times \frac{|0.01 - 0.05|}{10} \]
\[ |e| = 3 \times \frac{0.04}{10} \]
\[ |e| = \frac{0.12}{10} = 0.012 \text{ V} \]
To convert to millivolts (mV):
\[ |e| = 0.012 \times 10^{3} \text{ mV} = 12 \text{ mV} \] Step 4: Final Answer:
The induced e.m.f. is \(12 \text{ mV}\).