Question

A varying magnetic flux linking a coil is \(\Phi =xt^{2}\). If at time \(t=3\) s, the e.m.f. induced is \(9\)V then the magnitude of \(x\) is (in \(\text{Wb}/\text{s}^{2}\))

• A. \(1.5\)
• B. \(0.66\)
• C. \(0.33\)
• D. \(0.75\)

Hint:

Induced e.m.f. is the magnitude of the rate of change of flux. Differentiate and put t = 3 s.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
By Faraday's law, \(|e|=\left|\dfrac{d\Phi}{dt}\right|\).

Step 2: Differentiate
\(\Phi=xt^{2}\), so \(e=2xt\).

Step 3: Substitute
At \(t=3\) s, \(e=6x=9\) V.

Step 4: Solve
\(x=\dfrac{9}{6}=1.5\ \text{Wb/s}^{2}\).

Final Answer:
The value of x is 1.5 Wb per second squared. \[ \boxed{\text{(A) }1.5} \]