Figure shows a rectangular conducting loop PQRS in which arm RS o… - Vidyadip

Question

Figure shows a rectangular conducting loop $\text{PQRS}$ in which arm $RS$ of length I is movable. The loop is kept in a uniform magnetic field $B$ directed downward perpendicular to the plane of the loop. The arm $RS$ is moved with a uniform speed $v$.

Deduce an expression for
$i$. the emf induced across the arm $RS$
$ii.$ the external force required to move the arm and
$iii.$ the power dissipated as heat.

✓

Answer

$i.$ Let $RS$ moves with speed $v$ rightward and also $R S$ is at distances $x_1$ and $x_2$ from $PQ$ at instants $t_1$ and $t_2$, respectively.
Change in flux $, d \phi=\phi_2-\phi_1=B l\left(x_2-x_1\right)\left[\because \text { magnetic flux, } \phi=\vec{B} \cdot \vec{A}=B A \cos 0^0=B l x\right]$
$\Rightarrow d \phi=B l d x$
$ \Rightarrow \frac{d \phi}{d t}=B l \frac{d x}{d t}=B l v \quad\left[\because v=\frac{d x}{d t}\right]$
If resistance of loop is $R,$ then $I=\frac{v B I}{R}$
$ ii.$ Magnetic force $=B I l \sin 90^{\circ}$
$=\left(\frac{v B l}{R}\right) B l=\frac{v B^2 l^2}{R}$
Now, External force must be equal to magnetic force
$\therefore$ External force $=\frac{v B^2 l^2}{R}$
$\text { iii. As, } P=I^2 R=\left(\frac{ v B}{R}\right)^2 \times R=\frac{ v ^2 B^2 l^2}{R^2} \times R$
$\therefore P=\frac{ v ^2 B^2 l^2}{R}$

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