Question

Magnetic flux linked with the coil in weber is given by the equation \(\Phi = 5t^2+6t+11\). The e.m.f. induced in the coil in the \(5^{\text{th}}\) second will be

• A. \(10\) V
• B. \(51\) V
• C. \(102\) V
• D. \(166\) V

Hint:

The e.m.f. in the 5th second is the average rate of change of flux between t = 4 s and t = 5 s.

Answer 1

The Correct Option is B

Step 1: Faraday's Law:
The induced e.m.f. is the rate of change of flux. For the 5th second, which runs from \(t=4\) s to \(t=5\) s, the e.m.f. is the average rate over this second:
\[ \varepsilon=\left|\frac{\Phi(5)-\Phi(4)}{5-4}\right| \]

Step 2: Compute the Flux Values:
\(\Phi(5)=5(25)+6(5)+11=125+30+11=166\) Wb.
\(\Phi(4)=5(16)+6(4)+11=80+24+11=115\) Wb.

Step 3: Average e.m.f.:
\[ \varepsilon=\frac{166-115}{1}=51\ \text{V} \]

Step 4: Check the Other Options:
10 V is \(d\Phi/dt\) at \(t=1\) s minus 6, which is not meaningful. 102 V and 166 V do not match any rate of change of this flux over the 5th second.

Final Answer:
The e.m.f. induced in the 5th second is 51 V, option (B). \[ \boxed{\text{(B) } 51\ \text{V}} \]