Motional EMI (Induced Parameter) Questions in English

(B) The induced $EMF$ $(\varepsilon)$ in a conductor moving through a magnetic field is given by $\varepsilon = Bv\ell$.
First, calculate the velocity $(v)$ of the wire after falling a distance $(h = 200 \ m)$ under gravity using the equation $v^2 = u^2 + 2gh$. Since the initial velocity $(u = 0)$, $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10} \ m/s$.
The magnetic field $(B)$ is $0.5 \ Gauss = 0.5 \times 10^{-4} \ T$.
The length of the wire $(\ell)$ is $20 \ m$.
Substituting these values into the $EMF$ formula: $\varepsilon = (0.5 \times 10^{-4} \ T) \times (20\sqrt{10} \ m/s) \times (20 \ m)$.
$\varepsilon = 20\sqrt{10} \times 10^{-4} \times 10 = 20\sqrt{10} \times 10^{-3} \ V$.
Since $1 \ V = 1000 \ mV$, the induced $EMF$ is $20\sqrt{10} \ mV$.

(D) When the rod moves with a terminal velocity $v$,the induced electromotive force $(EMF)$ in the rod is $e = Bvl$.
Since the rod is part of a closed circuit with resistance $R$,the induced current is $i = \frac{e}{R} = \frac{Bvl}{R}$.
The magnetic force acting on the rod is $F_m = ilB = (\frac{Bvl}{R})lB = \frac{B^{2}l^{2}v}{R}$,which acts upwards.
At terminal velocity,the gravitational force $mg$ is balanced by the magnetic force $F_m$.
Therefore,$mg = F_m = \frac{B^{2}l^{2}v}{R}$.
Solving for $v$,we get $v = \frac{mgR}{B^{2}l^{2}}$.

(D) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B l v$.
Given: $B = 0.10 \ T$,$l = 1 \ m$,$v = 1.5 \ m/s$,and $R = 2 \ \Omega$.
The induced current in the circuit is $I = \frac{\varepsilon}{R} = \frac{B l v}{R}$.
The magnetic force acting on the rod is $F_B = I l B = \left( \frac{B l v}{R} \right) l B = \frac{B^2 l^2 v}{R}$.
To move the rod with a constant speed,the external force $F_{ext}$ must be equal and opposite to the magnetic force $F_B$.
$F_{ext} = F_B = \frac{B^2 l^2 v}{R}$.
Substituting the values:
$F_{ext} = \frac{(0.1)^2 \times (1)^2 \times 1.5}{2} = \frac{0.01 \times 1 \times 1.5}{2} = \frac{0.015}{2} = 0.0075 \ N$.
$F_{ext} = 7.5 \times 10^{-3} \ N$.

(A) The motional emf $d\varepsilon$ induced in a small element $dr$ at distance $r$ is $d\varepsilon = (v) B(r) dr$,where $v = \omega r$.
Thus,$d\varepsilon = (\omega r) (B_0 e^{-\lambda r}) dr$.
Integrating from $r=0$ to $L$: $\varepsilon = \int_0^L \omega B_0 r e^{-\lambda r} dr$.
Using integration by parts $\int r e^{-\lambda r} dr = -\frac{r}{\lambda} e^{-\lambda r} - \frac{1}{\lambda^2} e^{-\lambda r}$.
Evaluating from $0$ to $L$: $\varepsilon = \omega B_0 [(-\frac{L}{\lambda} e^{-\lambda L} - \frac{1}{\lambda^2} e^{-\lambda L}) - (0 - \frac{1}{\lambda^2})]$.
Simplifying the expression: $\varepsilon = B_0 \omega [\frac{1}{\lambda^2} - e^{-\lambda L} (\frac{1}{\lambda^2} + \frac{L}{\lambda})]$.

(A) The motional electromotive force (emf) induced in a conductor moving through a magnetic field is given by the formula $\varepsilon = B l v$,where $B$ is the magnetic field strength,$l$ is the length of the conductor moving perpendicular to the field,and $v$ is the velocity of the conductor.
Given values are: $B = 0.3 \text{ T}$,$l = 3 \text{ cm} = 0.03 \text{ m}$ (since the velocity is normal to the shorter side,the length of the side cutting the field lines is $3 \text{ cm}$),and $v = 2 \text{ cm s}^{-1} = 0.02 \text{ m s}^{-1}$.
Substituting these values into the formula:
$\varepsilon = 0.3 \text{ T} \times 0.03 \text{ m} \times 0.02 \text{ m s}^{-1}$
$\varepsilon = 0.00018 \text{ V} = 1.8 \times 10^{-4} \text{ V}$.
Therefore,the correct option is $A$.