Question 138: Electromagnetic Induction

Question

A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T . The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is
$\_\_\_\_$ mV . (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

The mass $(\mathrm{m})$ of the bob is $10 \mathrm{~g}=0.01 \mathrm{~kg}$
The length of wire is $\mathrm{l}=10 \mathrm{~cm}=0.1 \mathrm{~m}$.
The magnetic field is uniform in the region with strength, $B=2 T$.
The pendulum is released from angle $	heta=60^{\circ}$
The conducting rod of length 1 rotating around a fixed pivot point with a constant angular velocity $\omega$. The rotation takes place in a uniform magnetic field $B$ that is perpendicular to the plane of rotation.
Imagine a small element of length $d r$ on the rod located at a distance $r$ from the pivot point (axis of rotation).
The linear velocity v of this small element is related to the angular velocity $\omega$ by the relation:
$$ \mathrm{v}=\mathrm{r} \omega $$
The induced EMF ( $\mathrm{d} \varepsilon$ ) in a small conductor of length dr moving with velocity v perpendicular to a magnetic field $B$ is given by:
$$ \mathrm{d} \varepsilon=\mathrm{B} \cdot \mathrm{v} \cdot \mathrm{dr} $$
However, unlike a rod moving in a straight line, a rotating rod has different velocities at different points. The part near the pivot moves slowly, while the tip moves fastest as $\mathrm{v}=\mathrm{r} \omega \Rightarrow \mathrm{v} \propto \mathrm{r}$
To find the total EMF ( $\varepsilon$ ) across the entire rod, we integrate the small EMFs from the pivot ( $r=0$ ) to the tip of the $\operatorname{rod}(\mathrm{r}=\mathrm{l})$.
$$ \varepsilon=\int_0^{\varepsilon} \mathrm{d} \varepsilon=\int_0^{\mathrm{r}} \mathrm{~B} \cdot \mathrm{v} \cdot \mathrm{dr}=\int_0^{\mathrm{r}} \mathrm{~B} \cdot(\omega \mathrm{r}) \cdot \mathrm{dr} $$
$$\Rightarrow $$ $\varepsilon=\mathrm{B} \omega \int_0^1 \mathrm{rdr}$
$$\Rightarrow $$ $\varepsilon=\mathrm{B} \omega\left[\frac{\mathrm{r}^2}{2}ight]_0^1$
$$\Rightarrow $$ $\varepsilon=\mathrm{B} \omega\left(\frac{\mathrm{l}^2}{2}-\frac{0^2}{2}ight)$
$$\Rightarrow $$ $\varepsilon=\mathrm{B} \omega\left(\frac{\mathrm{l}^2}{2}ight)$
$$\Rightarrow $$ $$ \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{l}^2 $$
To find the maximum induced EMF, we need to find the point where the angular velocity $\omega$ (and consequently the linear speed v ) is maximum. For a pendulum, the speed is highest at the lowest point of its swing (the equilibrium position).
If the pendulum is released from an angle $	heta$, then the lost in height is,
$\mathrm{h}=\mathrm{l}(1-\cos 	heta)$
$$\Rightarrow $$ $\mathrm{h}=0.1 	imes\left(1-\cos 60^{\circ}ight)$
$$\Rightarrow $$ $\mathrm{h}=0.1 	imes(1-0.5)=0.05 \mathrm{~m}$
Using Conservation of Energy, when the pendulum is released from an angle $	heta$, it loses potential energy and gains kinetic energy as it swings down.
$$ \mathrm{mg} \Delta \mathrm{~h}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\max }^2 $$
$$\Rightarrow $$ $\mathrm{v}_{\max }=\sqrt{2 \mathrm{gh}}$
$$\Rightarrow $$ $\mathrm{v}_{\max }=\sqrt{2 	imes 10 	imes 0.05}$
$$\Rightarrow $$ $$ \mathrm{v}_{\max }=\sqrt{1}=1 \mathrm{~m} / \mathrm{s} $$
Using the relationship between linear velocity (v) and angular velocity ( $\omega$ ) is $\mathrm{v}=\omega \mathrm{l}$.
$$ \omega_{\max }=\frac{v_{\max }}{l} $$
$$\Rightarrow $$ $$ \omega_{\max }=\frac{1}{0.1}=10 \mathrm{rad} / \mathrm{s} $$
Substituting the values into the rotational EMF formula :
$$ \mathrm{E}_{\max }=\frac{1}{2} \mathrm{~B} \omega_{\max } \mathrm{l}^2 $$
$$\Rightarrow $$ $\mathrm{E}_{\max }=\frac{1}{2} 	imes 2 	imes 10 	imes(0.1)^2$
$$\Rightarrow $$ $\mathrm{E}_{\max }=1 	imes 10 	imes 0.01$
$$\Rightarrow $$ $\mathrm{E}_{\max }=0.1 \mathrm{~V}$
$$\Rightarrow $$ $\mathrm{E}_{\max }=0.1 	imes 1000 \mathrm{mV}$
$$\Rightarrow $$ $$ \mathrm{E}_{\max }=100 \mathrm{mV} $$
Therefore, the maximum induced EMF is 100 mV .
Hence, the correct answer is $\mathbf{1 0 0}$.

Electromagnetic Induction

Integer / Numeric Answer Type