Question

A$10cm\times 10cm$square loop lies in the $xy$-plane. The magnetic field in this region of space is $B=\left(0.30t\widehat{ı}+0.50t^2\widehat{k}\right)\mathrm{T}$, where $t$is in $s$. What is the $emf$induced in the loop at $\left(a\right)t=0.5s$and $\left(b\right)t=1.0s.$

Short answer

Part $\left(a\right)$

$\left(a\right)$The induced $emf$in the loop is$\epsilon =0.005\mathrm{V}.$

Part $\left(b\right)$

$\left(b\right)$The induced $emf$in the loops is$\epsilon =0.01\mathrm{V}.$

Step-by-step solution

Step: 1 Faraday's Law of Induction:

The induced emf in the loop is equivalent to the weight of the rate of change of the magnetic flux through the loop, according to Faraday's law of induction. The derivative of the magnetic flux with respect to time equals this rate of change, thus

$\epsilon =\left|\frac{d\Phi }{dt}\right|$

Now that the loop is in the plane, the loop's normal is the $z$-axis, this means that the plane's normal unit vector. As an outcome, only the component along contributes to the transit across the loop:

$\Phi =B_{z}A=a^{2}0.50t^{2}T.$

Step: 2 Derivative Portion:

where $A=a^2$is the area of the loop and $a=10\mathrm{cm}=0.1\mathrm{m}$is its side.

Taking the derivative we obtain

$$\begin{gathered} \frac{d\Phi }{dt}=\frac{d}{dt}\left(a^{2}0.50t^2\mathrm{T}\right) \\ \frac{d\Phi }{dt}=0.50a^2·2t\mathrm{T} \\ \frac{\mathrm{d\Phi }}{\mathrm{dt}}=1.0a^{2}t\mathrm{T}. \end{gathered}$$

Returning this into equation we get

$\epsilon =1.0a^{2}t\mathrm{T}.$

Step: 3 Finding the values: (part a and part b)

(a) Set $t=0.5\mathrm{s}$into the expression for $\epsilon$to get

$\epsilon =0.005\mathrm{V}.$

(b) Set $t=1.0\mathrm{s}$into the expression for $\epsilon$to get

$\epsilon =0.01\mathrm{V}.$