Question

The magnetic flux threading a metal ring varies with time $t$according to ${\Phi }_B=3(a{t}^3-b{t}^2)T{m}^2$with $a=2.00s^{-3}$and $b=6.00s^{-2}$The resistance of the ring is $3.0\Omega$ Determine the maximum current induced in the ring during the interval from $t=0$ to $t=2.0s.$

• A. $12A$
• B. $3A$
• C. $6A$
• D. $8A$

Answer 1

The correct option is C

$6A$

Step 1: Given data:

${\Phi }_B=3(a{t}^3-b{t}^2)$

Resistance of the ring $3.0\Omega$

$a=2.00s^{-3}$

$b=6.00s^{-2}$

Step 2: Finding the induced emf:

Induced emf $\left|e\right|=\frac{d{\varphi }_B}{dt}=\frac{d}{dt}\left[3\left(a{t}^3-b{t}^2\right)\right]=9a{t}^2-6bt$

Step 3: Find the induced current:

Induced current,

$i=\frac{\left|e\right|}{R}=\frac{9a{t}^2-6bt}{3}=3a{t}^2-2bt$

Step 4: Finding the time for maximum current:

For the current to be maximum

$\left(\frac{di}{dt}\right)=0$

$6at-2b=0$,$t=\frac{b}{3a}$, current is maximum.

Step 5: Finding the maximum current.

$$\begin{gathered} i_{max}=3a{\left(\frac{b}{3a}\right)}^2-2b\left(\frac{b}{3a}\right) \\ {i}_{max}=\frac{b^2}{3a}-\frac{2b^2}{3a}=\frac{b^2}{3a} \end{gathered}$$

Substituting the given values of $a$and $b$ we have,

$i_{max}=\frac{{\left(6\right)}^2}{2\times 3}=6.0A$

Hence, option$C$ is the correct answer.