Lenz’s Law

LENZ'S LAW

According to Lenz's law, induced emf in a circuit opposes the cause due to which it was induced.

Consider the following examples.

(a) Suppose that the north-pole of a bar magnet is moved towards a conducting wire loop as shown in the figure. Due to a change in the magnetic flux associated with the loop, a current is induced. Due to induced current, a magnetic field is induced and this magnetic field opposes the motion of bar magnet. The direction of the induced current can be deduced by the following argument: the north pole is moving towards the loop; therefore to oppose the motion of the bar magnet only a north pole will be induced on that face of the loop which faces the magnet.

(b) A rectangular loop ABCD is being pulled out of the magnetic field which is directed into the plane of the paper. Perpendicular to the plane of the paper. As the loop is dragged out of the field, the flux associated with the loop decreases. The induced current flows in the loop in a sense so as to oppose the decrease in this flux. For this to happen the magnetic field due to the induced current in the loop must be directed into the plane of the paper. Thus the current in the loop must flow be clockwise.

Illustration 1: Magnetic field is increasing into the page with time when a conducting loop of definite radius is placed on the plane of the paper. The find the direction of current in the loop.

Solution: As the flux is increasing inside then the current in the loop will be such that it will be opposing the increase in magnetic field, i.e., the induced current in the loop will create such a magnetic field which is directed out ward.

Thus the direction of current will be anticlockwise.

Illustration 2: A square loop ACDE of area 20cm² and resistance 5$\Omega$ is rotated in a magnetic field $\overrightarrow{B}$ = 2T through 180°

(a) in 0.01 s and

(b) in 0.02 s

Find the magnitude of e, i and $\Delta$q in both the cases.

Solution: Let us take the area vector $\overrightarrow{S}$ perpendicular to plane of loop inwards. So initially, $\overrightarrow{S}$$\uparrow \uparrow$$\overrightarrow{B}$ and when it is rotated by 180°, $\overrightarrow{S}$$\uparrow \downarrow$$\overrightarrow{B}$ Hence, initial flux passing through the loop,

$\varphi$_i = BS cos 0° = (2) (20 x 10^-4) (1) = 4.0 x 10^-3 Wb

Flux passing through the loop when it is rotated by 180°,

$\varphi$_f = BS cos 180° = (2) (20 x 10^-4) (-1) = - 4.0 x 10^-3 Wb

Therefore, change in flux,

$\Delta$$\varphi$_B = $\varphi$_f - $\varphi$_i = - 8.0 x10^-3 Wb

(a) Given $\Delta$t = 0.01 s, R=5 $\Omega$

$\therefore \left|e\right|=\left|-\frac{\Delta {\varphi }_B}{\Delta t}\right|=\frac{8.0\times {10}^{-3}}{0.01}=0.8volt$

$i=\frac{\left|e\right|}{R}=\frac{0.8}{5}=0.16A$

and $\Delta$q = i$\Delta$t = 0.16 x 0.01 = 1.6 x 10^-3 C

(b) $\Delta$t = 0.02 s

$\therefore \left|e\right|=\left|-\frac{\Delta {\varphi }_B}{\Delta t}\right|=\frac{8.0\times {10}^{-3}}{0.02}=0.4volt$

$i=\frac{\left|e\right|}{R}=\frac{0.4}{5}=0.08A$

and $\Delta$q = i$\Delta$t = (0.08) (0.02)

= 1.6 x 10^-3 C

Note: Time interval $\Delta$t in part (b) is two limes the time interval in part (a), so e and i are half while $\Delta$q is same.