Question

In the figure shown, the magnitude of magnetic force on the wire $ABC$ in the given uniform magnetic field will be:
$(B=2\text{Tesla})$

• A. $4(3+2\pi )\text{N}$
• B. $20\text{N}$
• C. $30\text{N}$
• D. $40\text{N}$

Answer 1

The correct option is B $20\text{N}$

The given wire consists of two sections, one is straight and other one is semicircle. For circular portion, the $\overrightarrow{L}$ is along $BC$ as per the direction of current.


In the above diagram, the displacement length vector will be along $\text{AC}$.

Thus, we can replace the whole wire $ABC$, with a straight wire $\text{AC}$, carrying a $2\text{A}$ current.

From the geometry,

$AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{3^2+4^2}=5\text{m}$

Now, ${\overrightarrow{F}}_m=I(\overrightarrow{L}\times \overrightarrow{B})$

$\because \overrightarrow{B}$is perpendicular to the plane of $\text{AC}$.

$F_m=BIL\sin {90}^{\circ }=BI\text{(AC)}$

substituting the values we get,

$F_m=2\times 2\times 5=20\text{N}$

The direction of magnetic force is perpendicular to $\text{AC}$, as shown in figure. This can be found out using $\overrightarrow{L}\times \overrightarrow{B}$.

Therefore, option $\left(B\right)$ is right.

Why this question? Tip: In such problems, focus on identifying the effective length vector ${\overrightarrow{L}}_{eff}$, which can be then used in the relation $\overrightarrow{F}=I(\overrightarrow{L_{eff}}\times \overrightarrow{B})$ to solve problem very easily.