Question

he magnetic flux through the loop shown in the figure belowincreases according to the relation Φ_B =6.0t² + 6.4t, where Φ_B is inmilliwebers and t is in seconds.

(a) What is the magnitude of the emf induced in the loopwhen t = 2.9 s?
mV

(b) What is the direction of the current through R?

leftright    insufficientinformation

Answer 1

Concepts and reason

This question is based upon the concept of the faraday law.

Initially, calculate the magnitude of the ${\rm{emf}}$ induced in the loop by using faraday law. Later, in the other part, calculate the direction of the current through the resistor.

Fundamentals

The expression for induced ${\rm{emf}}$ is as follows,

$\varepsilon = - N\frac{{d\Phi }}{{dt}}$

Here, $d\Phi$ change in flux and $dt$ change in time and $N$ is the number of turns.

(a)

The magnetic flux through the loop is,

${\Phi _B} = 6{t^2} + 6.4t$

The expression for induced ${\rm{emf}}$ is as follows,

$\varepsilon = - N\frac{{d{\Phi _B}}}{{dt}}$

Substitute $1$ for $N$ and $\left( {6{t^2} + 6.4t} \right) {\rm{milliweber}}$ for ${\Phi _B}$ in above equation.

$\begin{array}{c}\\\varepsilon = - 1\frac{d}{{dt}}\left( {6{t^2} + 6.4t} \right)\\\\ = - \left( {12t + 6.4} \right)\\\end{array}$

Substitute $2.9{\rm{ s}}$ for $t$ in the above equation.

$\begin{array}{c}\\\varepsilon = - \left( {12\left( {2.9{\rm{ s}}} \right) + 6.4} \right)\\\\ = \left( { - 34.8{\rm{ s }} - {\rm{6}}{\rm{.4}}} \right)\\\\{\rm{ = }} - {\rm{41}}{\rm{.2}} {\rm{mV}}\\\end{array}$

The magnitude of the induced ${\rm{emf}}$ in the loop is ${\rm{41}}{\rm{.2 mV}}$ .

(b)

From Lenz’s law induced magnetic field will oppose the change in magnetic flux. Using right hand rule, the current must flow clockwise to produce the magnetic field into the page. Magnetic flux is increase and out of the page.

Ans: Part a

The magnitude of the ${\rm{emf}}$ induced in the loop is ${\rm{41}}{\rm{.2 mV}}$ .

Part b

The direction of the current through the resistor is left.