Question

Two parallel, conducting rails $1$ and $2$ are kept perpendicular to a uniform magnetic field $\left(B\right)$. The rails are at seperation $l$ and are joined at their ends by a resistance $R$. A conducting bar $AB$ is placed on the rails making ${60}^{\circ }$ with them. The bar is pulled with a velocity $v$ parallel to the rails. Find the current in the resistance $R$.

• A. $\frac{\text{V}Bl}{R}$
• B. $\frac{2B\text{V}l}{R}$
• C. $\frac{B\text{V}l}{2R}$
• D. $\frac{B\text{V}l}{4R}$

Answer 1

The correct option is A $\frac{\text{V}Bl}{R}$

The motional emf produced due to velocity $v$ of rod is given by
$\epsilon =\text{V}B{l}_{\perp }...\left(i\right)$
where $l_{\perp }=\text{length of rod perpendicular to its velocity}$
$\Rightarrow$From the figure rod is making angle ${60}^{\circ }$ with horizontal (Direction of velocity of rod), hence
$\therefore l_{\perp }=AB\sin {60}^{\circ }=l...\left(ii\right)$

From Eq $\left(i\right)\&\left(ii\right)$:
$\epsilon =\text{V}B\left(AB\right)\sin {60}^{\circ }$
$\epsilon =\text{V}Bl$
Now, current in circuit is,
$i=\frac{\text{emf}}{R}$
$\therefore i=\frac{\text{V}Bl}{R}$
Current through resistance will be same as current in circuit.