Question

A rectangle loop $ABCD$ is placed near on infinite length current carrying wire. Magnetic force on loop is

• A. $1.25\times {10}^{-4}N,Attraction$
• B. $1.25\times {10}^{-4}N,Repulsion$
• C. $1.25\times {10}^{-5}N,Repulsion$
• D. $1.25\times {10}^{-5}N,Attraction$

Answer 1

Answer: (A)

$1.25\times {10}^{-4}N,Attraction$

Given: Current in infinite wire:25A, side of rectangular loop=25 cm and 10 cm , current in circular loop =15A

Solution: From the figure we come to know that the sides of AB and CD are symmetrical and having opposite direction of current .So, net force on these wires will cancel out .

Now net force acting on the rectangular loop will be the same as that of in side AD and BC.

As we know that the formula of magnetic force of infinite wire is given by:

$B=\frac{{\mu }_{0}i}{2\pi r}$

So ,magnetic force of infinite wire at AD ,$B=\frac{{\mu }_0\times 25}{2\pi \times 0.05}\otimes$

Remember here that r=0.05

magnetic force of infinite wire at BC ,$B=\frac{{\mu }_0\times 25}{2\pi \times 0.30}\otimes$

Remember here that r=0.05+0.25 =0.30

As we know that the formula of force of current carrying wire:

$F=i(\overrightarrow{l}\times \overrightarrow{B})$

Force on AD$=15\times 0.1\times B_{AD}\left(left\right)$

Force on BC= $15\times 0.1\times B_{BC}\left(right\right)$

Net force=Force on AD-Force on BC

$F=15\times 0.1[B_{AD}-B_{BC}]$

$F=15\times 0.1\times \frac{{\mu }_0\times 25}{2\pi }\times (\frac{1}{0.05}-\frac{1}{0.3})$

$F=1.25\times {10}^{-4}N$ [towards left i.e. attraction]

Hence the correct answer is A