Question

A coil consists of $200$ turns of wire having a total resistance of $2.0\Omega$.Each turn is a square of side $18\,cm$and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to $0.5\,T$ in $0.80\,s$. What is the magnitude of induced emf and current in the coil while the field is changing?

Answer 1

From Lenz’s laws we know that change of magnetic field with time induce a voltage across a current loop kept in the magnetic field.A magnetic field is a vector field in the neighbourhood of a magnet, electric current, or changing electric field, in which magnetic forces are observable. A magnetic field is produced by moving electric charges and intrinsic magnetic moments of elementary particles associated with a fundamental quantum property known as the spin.

Formula used:
The induced E.M.F across the loop is given by,
$\xi = - \dfrac{{d\varphi }}{{dt}}$
Where $\varphi$ is the flux through the loop, $\xi$ is the emf induced.
The flux passing through a square loop of $N$ turns is kept perpendicular to a uniform magnetic field $B$ is given by,
$\varphi = NB{A^2}$
where $A$ is the area of the square loop.
The current flowing through a circuit by Ohm’s law is given by,
$I = \dfrac{V}{R}$
where $V$ is the voltage drop across the circuit $R$ is the resistance of the circuit.

Complete step by step answer:
We have here a square loop of $200$ turns of wire having a total resistance of $2.0\Omega$ is kept in a uniform magnetic field changing linearly from 0 to $0.5\,T$ in $0.80\,s$. Now, we know from Lenz’s law the emf induced in a loop is given by the change of flux in the loop.The induced emf across the loop is given by,
$\xi = - \dfrac{{d\varphi }}{{dt}}$
Here, the change in flux is equal to,
$\Delta \varphi = N[B(0) - B(0.8)]{A^2}$

Putting the values we have,
$\Delta \varphi = 200 \times [0 - 0.5] \times {(0.18)^2}$
Now change in time is equal to,
$\Delta t = 0.8s$
Hence, the induced emf will be,
$\xi = - \dfrac{{\Delta \varphi }}{{\Delta t}}$
$\Rightarrow \xi = - \dfrac{{200 \times [0 - 0.5] \times {{(0.18)}^2}}}{{0.8}}$
Upon simplifying we will get,
$\xi = 4.05\,V$
Now, from Ohm’s law we know the current flowing through a circuit is given by, $I = \dfrac{V}{R}$
Where $V$is the voltage drop across the circuit $R$is the resistance of the circuit.
Hence, current through the loop will be,
$i = \dfrac{\xi }{R}$
$\Rightarrow i = \dfrac{{4.05}}{2}$
$\therefore i = 2.025$

So, current through the loop will be $2.025\,A$.

Note: If the loop is kept horizontal to the field applied the flux through the loop will be automatically zero. In that case no current will flow through the circuit. To solve this type of problem just recall Lenz's law to find the induced voltage in the loop.