Physics > Electromagnetic Induction > 4.0 Len’z Law:
Electromagnetic Induction
1.0 Introduction
2.0 Magnetic Flux
3.0 Experiments by Faraday and Henry
4.0 Len’z Law:
5.0 Induced Electric Field
6.0 Eddy Current:
7.0 Inductor and Inductance:
7.1 Inductors connected in series:
7.2 Inductors connected in parallel:
7.3 Mutually coupled inductors in parallel:
8.0 Growth and Decay of current in an LR circuit:
4.1 Motional EMF
7.2 Inductors connected in parallel:
7.3 Mutually coupled inductors in parallel:
(i) Consider a conducting rod is moving on a U shaped rail shown in figure. Magnetic field around the rail is $B$ in magnitude and direction is inside the plan. Rod is moving with velocity $v$. Consider friction between rail and rod is negligible. $HIJK$ is forming a closed circuit which area is changing as $ab$ is moving. At a particular time length between ad is $x$ and length of $JK$ is $l$. So magnetic flux enclosed by this loop is ${\Phi _m} = Bxl$ IMG
Induced emf is the rate of change of magnetic flux,
$\varepsilon = - \frac{{d{\Phi _m}}}{{dt}}$
=$\frac{{ - d\left( {Bxl} \right)}}{{dt}}$
=$ - Bl\frac{{dx}}{{dt}}$
$v = - \frac{{dx}}{{dt}}$and negative sign is because $x$ is decreasing with time
= $Blv$
Example: Figure shows a horizontal magnetic field which is uniform above the dotted line and zero below it. A long, rectangular, conducting loop of width $l$, mass m and resistance R is placed partly above and partly below the dotted line with the lower edge parallel to it. With what velocity should it be push downwards so that it may continue to fall without any acceleration?
Solution: Let the uniform velocity of fall be $v$. The emf is induced across the upper wire and its magnitude is $\varepsilon = vBl$. The current induced in the frame is
$$i = \frac{{vBl}}{R}$$
So that, the magnetic force on the upper arm is $F = ilB = \frac{{v{B^2}{l^2}}}{R}$. This force is in the upward direction. As the frame falls uniformly, this force should balance its weight. Thus,
$$mg = \frac{{v{B^2}{l^2}}}{R}$$
or, $$v = \frac{{mgR}}{{{B^2}{l^2}}}$$
(ii) A conducting rod $IJ$ is moving in Magnetic field of magnitude $B$ and direction inward the plane with velocity $v$. Assume the random velocity of electrons is zero on average, so we can consider that the electrons is moving with velocity $v$ . The Lorentz force on these electrons is $qvB$ in magnitude and this force is toward $J$ in direction. We can say that negative charge will accumulate at $J$ and positive charge will accumulate at $I$. An electric field $E$ will develop in conducting rod from $I$ to $J$. So because of this electric field a new force exerts on electrons and this force is in opposite direction of Lorentz force. When this force will equal to the Lorentz force then charge accumulation will stop. And at this condition,
${\vec F_B} = q\vec v \times \vec B$
${\vec F_E} = q\vec E$
$\left| {{{\vec F}_B}} \right| = \left| {{{\vec F}_E}} \right|$
$\left| {q\vec v \times \vec B} \right| = \left| {q\vec E} \right|$
$\left| {\vec v \times \vec B} \right| = \left| {\vec E} \right|$
$E = vB$
The potential difference between $IJ$ is,
$V = El$
The work done in moving a charge from $I$ to $J$ is,
$W = qvBl$
So produced EMF
$\varepsilon = \frac{W}{q}$
$=Blv$
Example: Figure shows a square loop having $50$ turns, an area of $62.5$ $ \times $ 10 and a resistance of $5$ $\Omega $. The magnetic field has a magnitude $B = 0.40 T$. Find the work done in pulling the loop out of the field, slowly and in $5.0$ sec.
Solution: The side of square is
$l = \sqrt {62.5 \times {{10}^{ - 3}}{m^2}} $
$ = 0.25m$
As it is uniformly pulled out in $5.0$ s, the speed of the loop is
$v = 0.5m{s^{ - 1}}$
The emf induced in left arm of the loop is
$\varepsilon = NvBl$
$ = 50 \times (0.05m{s^{ - 1}}) \times (0.40T) \times (0.25)$
$ = 0.25V$
The Current in the loop is
$i = \frac{{0.25V}}{{50\Omega }} = 5.0 \times {10^{ - 3}}A$
The force on the left arm due to magnetic field is
$F = NilB = 50 \times (5.0 \times {10^{ - 3}}A) \times (0.05m) \times (0.40T)$
$5.0 \times {10^{ - 3}}N$.
The force is toward left in the figure. To pull the loop uniformly, an external force of $5.0 \times {10^{ - 3}}N$ toward right must be applied. The work done by this force is
$W = (5.0 \times {10^{ - 3}}N) \times \left( {0.25m} \right) = 1.25 \times {10^{ - 3}}J$
