Question

A wire $\text{abc}$ is carrying current $i$. It is bent as shown in the figure, and is placed in a uniform magnetic field of magnetic induction $B$. The length of $\text{ab}$ is $l$ and $\angle abc={45}^{\circ }$. The ratio of magnetic force on $\text{ab}$ to that on $\text{bc}$ is:

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $1$
• D. $\frac{2}{3}$

Answer 1

The correct option is C $1$

Considering the triangle $\text{abc}$, the length of side $\text{bc}$ is given by:

$bc\cos {45}^{\circ }=ab$

or, $bc=\frac{ab}{\cos {45}^{\circ }}=l\sqrt{2}$

Now magnetic force on wire $\text{ab}$ is given by,

$F_{ab}=Bil\sin {90}^{\circ }=Bil$
$(\because$ angle between ${\overrightarrow{L}}_{ab}$ and $\overrightarrow{B}$ is ${90}^{\circ })$

Now, magnetic force on wire $\text{bc}$ is given by,

$F=B_i{l}_{bc}\sin \theta$

$(\because F=i({\overrightarrow{l}}_{BC}\times \overrightarrow{B})$ and $\theta ={45}^{\circ })$

$\Rightarrow F=Bi\left(l\sqrt{2}\right)\times \frac{1}{\sqrt{2}}$
or, $F_{bc}=B_{il}$

Thus, $\frac{F_{ab}}{F_{bc}}=\frac{Bil}{Bil}=1$

Hence,option $\left(C\right)$ is the correct answer.

Alternate solution: The equivalent length of $\text{bc}$ perpendicular to $\overrightarrow{B}$ is same as that of $\text{ab}$.

So, the forces on both the parts are equal.

Thus, $\frac{F_{ab}}{F_{bc}}=\frac{Bil}{Bil}=1$