Question
State Lenz's Law.
✓
Answer
Lenz's law: It is stated that the direction of induced e.m.f. is always in such direction that it opposes the change in magnetic flux.
$
e=\frac{d \varphi}{d t}
$
Consider a rectangular metal coil PQRS. Let ' $L$ ' be the length of the coil. It is placed in a partly magnetic field ' $B$ '. the direction of magnetic field is perpendicular to paper and into the paper. The ' $x$ ' part of the coil is in magnetic field at instant $t$. If the coil is moved towards right with a velocity $v=d x / d t$ with help of external agent like hand. The magnetic flux through the coil is $\Phi=B A=B L x \therefore \Phi=B L x$----(1) There is relative motion of a current through the coil. Let ' $i$ ' be current through the coil.
Three forces acts on the coil .
$F _1$ on conductor PL $\therefore F 1=$ Bix, vertically upward.
$F _2$ on conductor MS $\therefore F 2=$ Bix, vertically downward.
$F _3$ on conductor SP $\therefore F 3=$ BiL towards left. $F_1 \& F_2$ are equal and opposite and also in a same lines. They will cancel each other, $F_3$ is a resultant force. The external agent has to do work against this force.
$\therefore F 3=- Bil - ve$ sign indicates that force is opposite to $dx$.
If $dx$ is displacement in time $dt$, then work done $dw = F 3 dx$
$
\therefore dw =- \text { BiL } dx
$
This power is an electrical energy 'ei' where ' $e$ ' is an induced e.m.f.
$
\begin{aligned}
& \therefore e i=-\frac{B i l d x}{d t} \\
& \therefore e=-\frac{B L d x}{d t}
\end{aligned}
$
This power is an electrical energy 'ei' where ' $e$ ' is an induced e.m.f.
$
\begin{aligned}
& \therefore e i=-\frac{B i l d x}{d t} \\
& \therefore e=-\frac{B L d x}{d t}
\end{aligned}
$
$
\therefore e=-B L v
$
$\therefore e=-\frac{d}{d t}(B L x)$
$\therefore=e=-\frac{d \varphi}{d t}$ from eq (1)
$
e=\frac{d \varphi}{d t}
$
Consider a rectangular metal coil PQRS. Let ' $L$ ' be the length of the coil. It is placed in a partly magnetic field ' $B$ '. the direction of magnetic field is perpendicular to paper and into the paper. The ' $x$ ' part of the coil is in magnetic field at instant $t$. If the coil is moved towards right with a velocity $v=d x / d t$ with help of external agent like hand. The magnetic flux through the coil is $\Phi=B A=B L x \therefore \Phi=B L x$----(1) There is relative motion of a current through the coil. Let ' $i$ ' be current through the coil.
Three forces acts on the coil .
$F _1$ on conductor PL $\therefore F 1=$ Bix, vertically upward.
$F _2$ on conductor MS $\therefore F 2=$ Bix, vertically downward.
$F _3$ on conductor SP $\therefore F 3=$ BiL towards left. $F_1 \& F_2$ are equal and opposite and also in a same lines. They will cancel each other, $F_3$ is a resultant force. The external agent has to do work against this force.
$\therefore F 3=- Bil - ve$ sign indicates that force is opposite to $dx$.
If $dx$ is displacement in time $dt$, then work done $dw = F 3 dx$
$
\therefore dw =- \text { BiL } dx
$
This power is an electrical energy 'ei' where ' $e$ ' is an induced e.m.f.
$
\begin{aligned}
& \therefore e i=-\frac{B i l d x}{d t} \\
& \therefore e=-\frac{B L d x}{d t}
\end{aligned}
$
This power is an electrical energy 'ei' where ' $e$ ' is an induced e.m.f.
$
\begin{aligned}
& \therefore e i=-\frac{B i l d x}{d t} \\
& \therefore e=-\frac{B L d x}{d t}
\end{aligned}
$
$
\therefore e=-B L v
$
$\therefore e=-\frac{d}{d t}(B L x)$
$\therefore=e=-\frac{d \varphi}{d t}$ from eq (1)
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