# a conducting loop consisting of a half circle of radius r=0.2m and three straight sections the...

## Question:

a conducting loop consisting of a half circle of radius r=0.2m and three straight sections the half circle lies in a uniform magnetic field B that is directed out of the page the field magnitude is given by B=(4t{eq}^2 {/eq} + 2t + 3)T where t is in second. An ideal battery with emf=2v is connected to the loop. The resistance of the loop is 2 ohm.

a) what is the magnitude of the induced emf at t=10sec ?

b) what is the current in the loop at t=10sec?

## EMF:

EMF stands for the electro-motive force that is equivalent to the potential difference over the ends of the cell or a battery when there is no movement of charge. The EMF is the voltage produced by a battery or a cell per coulomb of charge crossing through it. The standardized unit of EMF is volts.

## Answer and Explanation:

Given data

• The radius of half circle is : {eq}r = 0.2\,{\rm{m}} {/eq}
• The ideal battery with emf is : {eq}V = {\rm{2}}\,{\rm{V}} {/eq}
• The resistance of the circuit is : {eq}R = 2\,{\rm{\Omega }} {/eq}
• The time given is : {eq}t = 10{\kern 1pt} \,{\rm{s}} {/eq}

(a)

The expression to calculate the e.m.f. induced at time equals to 10 s is ,

{eq}{V_{e.m.f.}} = \dfrac{{d\phi }}{{dt}}\,.....\,(I) {/eq}

Here, {eq}\dfrac{{d\phi }}{{dt}} {/eq} is the rate of change of magnetic flux.

Calculate the rate of change of magnetic flux ,

{eq}\dfrac{{d\phi }}{{dt}} = \dfrac{{d\left( {B \times A\cos 90} \right)}}{{dt}} {/eq}

Here, {eq}A {/eq} is the area ,calculate the area of half circle,

{eq}\begin{align*} A& = \dfrac{{\pi {r^2}}}{2}\\ A &= \dfrac{{\pi \times 0.2 \times 0.2}}{2}\\ A &= 0.0628\,{{\rm{m}}^{\rm{2}}} \end{align*} {/eq}

Substituting the value in the above expression

{eq}\begin{align*} \dfrac{{d\phi }}{{dt}}& = \dfrac{{d\left( {\left( {4{t^2} + 2t + 3} \right) \times 0.0628} \right)}}{{dt}}\\ \dfrac{{d\phi }}{{dt}} &= 0.0628\left( {8t + 2} \right) \end{align*} {/eq}

For time t=10s,

{eq}\begin{align*} \dfrac{{d\phi }}{{dt}}& = 0.0628\left( {8 \times 10 + 2} \right)\\ \dfrac{{d\phi }}{{dt}} &= 5.15\,{\rm{V}} \end{align*} {/eq}

Substituting the value in the expression (I).

{eq}{V_{e.m.f.}} = 5.15\,{\rm{V}} {/eq}

Thus, the magnitude of induced e.m.f. is equal to {eq}5.15\,{\rm{V}} {/eq}

(b)

The expression to calculate the current in the loop is,

{eq}I = \dfrac{{{V_{emf}} - V}}{R} {/eq}

Substituting the value in the above expression,

{eq}\begin{align*} I &= \dfrac{{5.15 - 2}}{2}\\ I &= 1.575\,{\rm{A}} \end{align*} {/eq}

Thus, the magnitude of current is {eq}1.575\,{\rm{A}} {/eq} .