Physics 3 Marks Question

Initial current passing = i

Hence initial emf = ir

Emf due to motion of ab = Blv

Net emf = ir - Blv

Net resistance = 2r

Hence current passing $=\frac{\text{ir}-\text{Blv}}{2\text{r}}$

1. The wires constitute 2 parallel emf.

$\therefore$ Net emf $=\text{Blv}=1\times4\times10^{-2}\times5\times10^{-2}=20\times10^{-4}$

Net resistance $=\frac{2\times2}{2+2}+19=20\Omega$

Net current $=\frac{20\times10^{-4}}{20}=0.1\text{mA}$

2. When both the wires move towards opposite directions then not emf = 0

$\therefore$ Net current = 0

1. For maximum emf, $\sin\theta=1$

$\therefore\text{e}=\text{BA}\omega$

$\Rightarrow\text{e}=0.010\times25\times10^{-4}\times80\times\frac{2\pi\times\pi}{60}$

$\Rightarrow\text{e}=0.66\times10^{-3}=6.66\times10^{-4}\text{V}$

2. The direction of the induced emf changes every instant. Thus, the average emf becomes zero.

3. The emf induced in the coil is $\text{e}=-\text{BA}\omega\sin\theta=-\text{BA}\omega\sin \ \omega\text{t}$

The average of the squares of emf induced is given by

$\text{e}_{\text{av}^2}=\frac{\int\limits_{0}^{\text{T}}\text{B}^2\text{A}^2\omega^2\sin^2\omega\text{t} \ \text{dt}}{\int\limits_{0}^{\text{T}}\text{dt}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2\int\limits_{0}^{\text{T}} \sin^2\omega\text{t} \ \text{dt}}{\int\limits_{0}^{\text{T}}\text{dt}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2\int\limits_{0}^{\text{T}} (1-\cos2\omega\text{t} )\ \text{dt}}{ \ 2\text{T}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2 }{ 2\text{T}}\Big[\frac{\text{t}-\sin2\omega\text{t}}{ \ 2\omega}\Big]_{0}^{\text{T}}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{\text{B}^2\text{A}^2\omega^2}{ \ 2\text{T}}\Big[\text{T}-\frac{\sin4\pi-\sin0}{2\omega}\Big]=\frac{\text{B}^2\text{A}^2\omega^2}{2}$

$\Rightarrow\text{e}_{\text{av}^2}=\frac{(6.66\times10^{-4})2}{2}$

$=22.1778\times10^{-8}\text{V}^2 \ \big[\because\text{BA}\omega=6.66\times10^{-4}\text{V}\big]$

$\Rightarrow\text{e}-{\text{av}^2}=2.2\times10^{-7}\text{V}^2$

1. The current through the wire is given by

$\text{i}=\frac{\text{E}-\text{Bvl}}{\text{r}}$

The direction of the current is from b to a.

2. The force acting on the wire at the given instant is given by

$\text{F}=\text{ilB}=\Big(\frac{\text{E}-\text{Bvl}}{\text{r}}\Big)$ towards right

3. The velocity of the wire attains a value such that it satisfies E = Bvl.

The net force on the wire becomes zero. Thus, the wire moves with a constant velocity v.

$\therefore\text{v}=\frac{\text{E}}{\text{Bl}}$

1. The perpendicular component i.e. a $\sin\theta$ is to be taken which is $\perp\text{r}$ to velocity.

So, $\text{I}=\text{a}\sin\theta \ 30^{\circ}=\frac{\text{a}}{2}$

Net ‘a’ charge $=4\times\frac{\text{a}}{2}=2\text{a}$

So, induced emf $=\text{B}\delta\text{l}=2\text{auB}$

2. Current $=\frac{\text{E}}{\text{R}}=\frac{2\text{auB}}{\text{R}}$

3. Total charge q, which flows through the sides of the frame, is given by

$\text{q}=\frac{\Delta\phi}{\text{R}}$

Here,

$\Delta\phi=$ Change in the flux

R = Resistance of the coil

$\therefore \ \text{q}=\frac{\Delta\phi}{\text{R}}$

$=\frac{\text{B}(\text{a}^2-0)}{\text{R}}$

$=\frac{\text{B}\text{a}^2}{\text{R}}$

We know

$\text{F}=\frac{\text{B}^2\text{a}^3\omega}{2\text{R}}=\text{il}\text{B}$

Component of mg along $\text{F}=\text{mg } \sin \theta$

Net force $=\frac{\text{B}^2\text{a}^3\omega}{2\text{R}}-\text{mg }\sin\theta$

$\text{B}=0.40\text{T},\omega=10 \ \text{rad}/',\text{r}=10\Omega$

$\text{r}=5\text{cm}=0.05\text{m}$

Considering a rod of length 0.05m affixed at the centre and rotating with the same $\omega$.

$\text{V}=\frac{\text{L}}{2}\times\omega=\frac{0.05}{2}\times10$

$\text{e}=\text{BLV}=0.40\times\frac{0.05}{2}\times10\times0.05=5\times10^{-3}\text{V}$

$\text{L}=\frac{\text{e}}{\text{R}}=\frac{5\times10^{-3}}{10}=0.5 \ \text{mA}$

It leaves from the centre.

1.  $\text{e}=50\text{V}, \ \text{T}=0.25\text{s}$

$\text{i}=\frac{\text{e}}{\text{R}}=2\text{A}, \text{H}=\text{i}^2\text{RT}$

$=4\times25\times0.25=25\text{J}$

2. During its restoration.

$\text{e}=50\text{V}, \ \text{T}=0.25\text{s}$

$\text{i}=\frac{\text{e}}{\text{R}}=2\text{A}, \text{H}=\text{i}^2\text{RT}=25\text{J}$

3. During its motion.

Since energy is a scalar quantity

Net thermal energy developed = 25J + 25J = 50J. 

1. The e.m.f. is highest between diameter $\perp\text{r}$ to the velocity. Because here length $\perp\text{r}$ to velocity is highest.

E_max = VB2R

2. The length perpendicular to velocity is lowest as the diameter is parallel to the velocity E_min = 0.

1. $\text{i}_{0} = \frac{\text{e}}{\text{R}}=\frac{4}{20},\tau=\frac{\text{L}}{\text{R}}=\frac{2}{20}=0.1 $

$ \text{i}=\text{i}_{0}(1-\text{e}^{\frac{-t}{\tau}})=\frac{4}{20}\Big(1-e^\frac{-0.2}{0.1}\Big) $

$= 0 .17\text{A}$

2. $ \frac{1}{2}\text{Li}^2 =\frac{1}{2} \times2\times(0.17)^2=0.0289=0.03\text{J}$

$\phi$ througth the coil

$\phi=\text{N.B.A}$

$\phi=\text{N.}\mu_0\text{ni}\pi\text{R}^2$

1. $\text{e}=\frac{\text{d}\phi}{\text{dt}}$

$=\frac{\text{d}}{\text{dt}}\big(\mu_0\text{n}\text{N}\pi\text{R}^2\text{i}\big)$

$=\mu_0\text{nN}\pi\text{R}^2\frac{\text{d}}{\text{dt}}(\text{i}_0\sin\omega\text{t})$

$=\mu_0\text{nN}\pi\text{R}^2\text{i}_0\cos\omega\text{t}\times\omega$

$\text{e}=\mu_0\text{nN}\pi\text{R}^2\text{i}_0\omega\cos\omega\text{t}$

2. $\text{E}=\text{M}\frac{\text{d}\text{I}}{\text{dt}}$

$\text{E}=\mu_0\text{nN}\pi\text{R}^2\frac{\text{d}\text{i}}{\text{dt}}$

$\text{M}=\mu_0\text{nN}\pi\text{R}^2$