Question

A long straight wire is carrying a current of $25A$. A square loop of side length $10cm$ carrying a current of $15A$, is placed $2cm$ from the wire as shown in figure. The force on the loop is

• A. $3.1\times {10}^{-2}$ towards the wire
• B. $3.1\times {10}^{-3}$ towards the wire
• C. $9.4\times {10}^{-4}$ towards the wire
• D. $9.4\times {10}^{-4}$ away from the wire

Answer 1

The correct option is B $3.1\times {10}^{-3}$ towards the wire

The wire carrying $25A$ current produces a magnetic field in its surrounding

the wires $AB$ and $CD$ are equal in length and carry current in opposite direction

Therefore two equal and opposite forces are produces and cancel out each other

$\Rightarrow |\overrightarrow{F_2}|=|\overrightarrow{F_4}|$ and opposite direction

Considering wire $AD$ and $BC$

Force between two current carrying wires

$F=\frac{\mu .I_1,I_2}{2\pi d}$ where, $d=$ distance $b/w$ the two wires

Wire $AD$

$F_1=\frac{{\mu }_o\left(25\right)\left(15\right)}{2\pi (2\times {10}^{-2})}$

$F_1=\frac{4\pi \times {10}^{-7}\times 25\times 15\times {10}^2}{2\pi \times 2}$

$F_1=25\times 15\times {10}^{-5}$

$F_1=375\times {10}^{-5}N$

Wire $BC$

$F_3=\frac{{\mu }_o\left(25\right)\left(15\right)}{2\pi (12\times {10}^{-2})}$

$F_3=\frac{4\pi \times {10}^{-7}\times 25\times 15\times {10}^2}{2\pi \times 12}$

$F_3=\frac{25\times 5\times {10}^{-5}}{2}$

$F_3=62.5\times {10}^{-5}N$

Thus net force will be

${\overrightarrow{F}}_{not}=(375-62.5)\times {10}^{-5}=312.5\times {10}^{-5}N$ (towards left)