Question

The adjoining figure shows a current carrying loop where the arc $PQRS$ is a three quarter circle and $PS$ is a straight line segment. A uniform magnetic field $B$ exists throughout the region with its direction being along the $-z$ axis. If it is given that the force on branch $PS$ is $F$ , then calculate the force on branch $PQR$ in terms of $F$ .

• A. $\sqrt{2}F$
• B. $\frac{F}{\sqrt{2}}$
• C. $\frac{\pi F}{\sqrt{2}}$
• D. $\sqrt{2}\pi F$

Answer 1

Answer: (A)

$POS$ is a right angled triangle. where
${\left(PS\right)}^2={\left(PO\right)}^2+{\left(OS\right)}^2$
$\Rightarrow PS=\sqrt{r^2+r^2}=\sqrt{2}r$
$\therefore$ Length of $PS$ will be equal to
$PS=\sqrt{2}r$
$F_{PS}=ilB=i\sqrt{2}rB=F(Given)$
Consider the vector length of the semicircular arc $PQR$ which is equal to the length $PR$ .
$PQR=2r$
$F_{PQR}=i(PR)B$
$\Rightarrow F_{PQR}=i.2r.B$
$\Rightarrow F_{PQR}=2irB$
$\Rightarrow F_{PQR}=\sqrt{2}F\left[\because F=\sqrt{2}irB\right]$