Answer to Question #192912 in Electric Circuits for hello

Question

A square current loop is placed in a uniform magnetic field of magnitude 0.65 T directed into the page. The loop’s side length changes from 20.0 cm to 6.0 cm in 0.50 s.

(a) What is the magnitude of the induced emf?

(b) If the loop has a resistance of 2.5 Ω, then what is the magnitude of the induced current?

Answer 1

Explanations & Calculations

• According to the definition for the induced EMF: $\small E=\Large\frac{d\phi}{dt}$ , when there is a change in magnetic flux, there is a voltage induced.
• We can expand the relationship further as follows

$\qquad\qquad \begin{aligned} \small E&=\small \frac{d(A\times\vec{B})}{dt}\cdots\cdots(\phi=area\times perpendicular field)\\ &=\small \vec{B}\cdot\frac{dA}{dt}\cdots\cdots(\because the\, field\,is\,a\,constant)\\ \end{aligned}$

• Then it is about evaluating the rate of change in the area of the loop.
• The area of the loop at the beginning

$\qquad\qquad \begin{aligned} \small A_1&=\small (0.2m )^2=0.04m^2\\ \end{aligned}$

• Final area

$\qquad\qquad \begin{aligned} \small A_2&=\small (0.06\,m)^2=0.0036\,m^2 \end{aligned}$

• Change accured

$\qquad\qquad \begin{aligned} \small \Delta A&=\small 0.0364\,m^2 \end{aligned}$

• Then the rate of change of area is

$\qquad\qquad \begin{aligned} \small \frac{dA}{dt}&=\small \frac{0.0364m^2}{0.50\,s}\\ &=\small 0.0728\,m^2s^{-1} \end{aligned}$

• Then the induced voltage is

$\qquad\qquad \begin{aligned} \small E&=\small 0.65T\times0.0728 m^2s^{-1}\\ &=\small \bold{0.047\,V} \end{aligned}$

2)

• Then we can imagine the situation as a battery of that E embedded in that loop in series with a resistor with a resistance equal to that of the loop.
• Then the flowing current is

$\qquad\qquad \begin{aligned} \small i&=\small \frac{E}{R}=\frac{0.047V}{2.5\Omega}\\ &=\small \bold{0.019 A\,(19\,mA)} \end{aligned}$