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 $A$.

(B) When the loop moves downwards with velocity '$v$' in a magnetic field '$B$',an induced electromotive force $(EMF)$ is generated across the width '$l$' of the loop, given by $\varepsilon = Blv$.
Since the loop has resistance '$R$',an induced current '$I$' flows through it, given by $I = \frac{\varepsilon}{R} = \frac{Blv}{R}$.
The magnetic force '$F$' acting on the horizontal segment of the loop in the magnetic field is $F = BIl = B(\frac{Blv}{R})l = \frac{B^2 l^2 v}{R}$.
This force acts upwards, opposing the gravitational force '$mg$'.
For the loop to fall without any acceleration, the net force must be zero, which means the magnetic force must balance the gravitational force:
$F = mg$
$\frac{B^2 l^2 v}{R} = mg$
Solving for '$v$':
$v = \frac{mgR}{B^2 l^2}$.

(B) The area of the square loop is $A = 25 \ cm^2 = 25 \times 10^{-4} \ m^2$. The side length of the loop is $l = \sqrt{A} = 5 \times 10^{-2} \ m$.
The magnetic field $B = 40 \ T$ and resistance $R = 10 \ \Omega$.
When the loop is pulled out of the magnetic field with velocity $v$, the induced electromotive force $(EMF)$ is $\varepsilon = Blv$.
The time taken to pull the loop out is $t = 1 \ s$. The velocity is $v = l/t = (5 \times 10^{-2} \ m) / (1 \ s) = 5 \times 10^{-2} \ m/s$.
The induced $EMF$ is $\varepsilon = 40 \times (5 \times 10^{-2}) \times (5 \times 10^{-2}) = 0.1 \ V$.
The power dissipated as heat is $P = \varepsilon^2 / R = (0.1)^2 / 10 = 0.01 / 10 = 10^{-3} \ W$.
The work done is equal to the heat dissipated: $W = P \times t = 10^{-3} \ W \times 1 \ s = 1.0 \times 10^{-3} \ J$.

(B) When a metal rod of length '$L$' moves with a constant velocity '$V$' in a magnetic field '$B$',an induced electromotive force $(EMF)$ is generated across the rod, given by $\varepsilon = BLV$.
Since the rod completes a circuit with total resistance '$R$',the induced current '$I$' flowing through the circuit is $I = \frac{\varepsilon}{R} = \frac{BLV}{R}$.
The magnetic force '$F_m$' acting on the current-carrying rod in the magnetic field is given by $F_m = I L B$.
Substituting the value of '$I$',we get $F_m = (\frac{BLV}{R}) L B = \frac{B^2 L^2 V}{R}$.
To move the rod at a constant speed, an external force '$F$' must be applied equal and opposite to the magnetic force. Therefore, the required force is $F = \frac{B^2 L^2 V}{R}$.

(A) The induced electromotive force (e.m.f.) $\varepsilon$ in a conductor moving perpendicular to a magnetic field is given by the formula: $\varepsilon = B \cdot l \cdot v$
Where:
$B = 0.8 \ Wb \ m^{-2}$ (Magnetic field intensity)
$l = 0.4 \ m$ (Length of the conductor)
$v = 7.0 \ ms^{-1}$ (Speed of the conductor)
Substituting the values:
$\varepsilon = 0.8 \times 0.4 \times 7.0$
$\varepsilon = 0.32 \times 7.0$
$\varepsilon = 2.24 \ V$
Thus, the induced e.m.f. across the conductor is $2.24 \ V$.

(B) The motional electromotive force $(EMF)$ induced across the wings of an aeroplane moving through a magnetic field is given by the formula: $\varepsilon = B \cdot l \cdot v$
Where:
$B = 3.6 \times 10^{-5} \ T$ (Magnetic field)
$l = 30 \ m$ (Wing span)
$v = 170 \ m/s$ (Velocity)
Substituting the values into the equation:
$\varepsilon = (3.6 \times 10^{-5}) \times 30 \times 170$
$\varepsilon = 3.6 \times 5100 \times 10^{-5}$
$\varepsilon = 18360 \times 10^{-5} \ V$
$\varepsilon = 0.1836 \ V$
Converting to millivolts $(mV)$:
$0.1836 \ V = 183.6 \ mV$
Therefore, the potential difference between the tips of the wings is $183.6 \ mV$.

(A) The motional emf induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $\varepsilon = Blv$, provided the velocity, magnetic field, and length are mutually perpendicular.
Step $1$: For the region $0 \le x < b$, the conductor is moving through a uniform magnetic field $B$. Thus, the induced emf is $\varepsilon = Blv$.
Step $2$: For the region $x > b$, the magnetic field $B = 0$. Therefore, the induced emf is $\varepsilon = B \cdot l \cdot v = 0 \cdot l \cdot v = 0$.
Conclusion: The induced emf is $Blv$ for $0 \le x < b$ and $0$ for $b \le x < 2b$.