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MODEL PAPER 7 — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsMODEL PAPER 73 Marks
Question
Figure shows a rectangular conducting loop 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.
Image

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.
$\begin{array}{l}\text { Change in flux, } d \phi=\phi_2-\phi_1=B l\left(x_2-x_1\right)\left[\because \text { magnetic flux, } \phi=\vec{B} \cdot \vec{A}=B A \cos 0^0=B l x\right] \\ 
\Rightarrow d \phi=B l d x \Rightarrow \quad \frac{d \phi}{d t}=B l \frac{d x}{d t}=B l v \quad\left[\because v=\frac{d x}{d t}\right]\end{array}$
If resistance of loop is R, then $I=\frac{v B I}{R}$
$\begin{array}{l}\text { ii. Magnetic force }=B I l \sin 90^{\circ} \\ \quad
=\left(\frac{v B l}{R}\right) B l=\frac{v B^2 l^2}{R}\end{array}$
Now, External force must be equal to magnetic force 
$\therefore$ External force $=\frac{v B^2 l^2}{R}$
$\begin{array}{l}\text { iii. As, } P=I^2 R=\left(\frac{ v B}{R}\right)^2 \times R=\frac{ v ^2 B^2 l^2}{R^2} \times R \\ \quad 
\therefore P=\frac{ v ^2 B^2 l^2}{R}\end{array}$

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