Write statement of Lenz's law of electromagnetism. A 2 m long straight vertical wire is falling perpendicularly to horizontal component at speed of 5 $m / s$ in the magnetic field of earth of $0.3 \times 10^{-4} T$. Calculate the magnitude of instantaneous value of induced emf across the ends of the wire.
Answer
Lenz's law : Polarity of induced electromotive force in the phenomenon of electromagnetic induction is such that it always opposes the cause which produces it means it opposes the change in magnetic flux due to which it is generated. $ E \propto-\frac{d \phi}{d t} \text { or } E=-\frac{k d \phi}{d t} $ Here, k is a constant of proportionality whose value is 1. Negative sign represents that induced emf always opposes change in flux. Given : $l=2 m$ $ \begin{aligned} B_{E} & =0.3 \times 10^{-4} T \\ E & =B / v \sin \theta \\ E & =B / v 90^{\circ} \end{aligned} $ On putting the values, $ \begin{aligned} E & =\left(0.3 \times 10^{-4}\right) \times 2 \times 5 \times 1 \quad \therefore \sin 90^{\circ}=1 \\ & =3 \times 10^{-4} \text { volt. } \end{aligned} $
What will be maximum mutual inductance between coils of self-inductance $L_1$ and $L_2$ ?
Answer
When a coil is wounded over the other coil, then due to complete pairing of two coils, the coefficient of mutual inductance is maximum and as a result magnitude of induced emf is also maximum. When the two coils are winded in any other way, coefficient of mutual inductance and induced emf do not have maximum values as flux in both coils is not completely associated. Two coils whose inductance are $L _1$ and $L _2$ and their mutual inductance is M . Then relation between $L _1, L_2$ and $M = K \sqrt{ L _1+ L _2}$ Where K is pairng coefficient of coils (solenoids). In ideal condition, $K =1$ Hence, maximum value of $M$ will be $\sqrt{L_1+L_2}$.
Radius of a disc of copper is 10 cm . It is rotating about an axis which is perpendicular to its plane and passing through its centre at the rate of $20 \pi rad /$ second. Uniform magnetic field of 0.2 T is working perpendicular to the disc. (i) Calculate potential difference between axis and circumference of the disc. (ii) If resistance of disc is 2 , then what is the magnitude of induced current?
Answer
Here, radius of disc $R =10 cm=0.10 m$, angular velocity of disc $\omega=20 \pi rad / second$, magnetic field $B=0.2$ T , resistance of disc $R =2 \Omega$. (i) Time taken by disc to complete one revolution crossing the magnetic field $\Delta t =2 \pi / \omega$ and area covered in completing one revolution is $\Delta A =\pi r^2$. Now developed potential difference between its axis and circumference due to change in magnetic flux associated with it. $V =|\varepsilon|=\frac{\Delta \phi}{\Delta t}=\frac{\Delta( BA )}{\Delta t }= B \cdot\left(\frac{\Delta A }{\Delta t }\right)$ $=\frac{ B \times \pi r ^2}{2 \pi / \omega}=\frac{1}{2} Br ^2 \omega$ On putting the values, $V =\frac{1}{2}(0.2)(0.10)^2 \times 20 \pi$ volt $=0.1 \times(0.10)^2 \times 20 \times 3.14 volt$ $=0.0628$ volt (ii) Induced current, $I =\frac{ V }{ R }=\frac{0.0628 Volt }{5 Ohm }$ $=0,0314$ ampere
A horizontal straight wire 10 m long extending from east to west is falling with a speed of $5.0 ms^{-1}$, at right angles to component of the earth's magnetic field $0.30 \times 10^{-4} Wbm ^{-2}$. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?
Answer
Given: $\quad l=10 m$ $B _{ E }=0.30 \times 10^{-4} Wb m ^{-2}$ $v=5.0 m / s ^{-1}$ (a) Let the value of instantaneous induced electromotive force $\varepsilon=$ ? On using formula $\varepsilon= B l v$ $\therefore \quad \varepsilon= B _{ E } l v$ $\because$ Is the horizontal component. On putting the values $=0.30 \times 10^{-4} \times 10 \times 5$ $=1.5 \times 10^{-3} V$ $=1.5 mV$ (b) Direction of induced emf will be from west to east which can be obtained by Fleming's right hand rule. (c) Since induced emf opposes the cause which produces it, hence it is situated from less to high end. Hence eastern end will be at higher electrical potential because emf works from west to east.