Question

A rectangular loop of sides 25 cm and 10 cm carrying current of 15 A is placed with its longer side parallel to a long straight conductor $2.0cm$ apart carrying a current of 25 A. What is the new force on the loop?

Answer 1

Hint: The net force is found by the difference of the repelling forces. The current flow along the shorter sides of the rectangular loop is equal and opposite, by symmetry they exert equal and opposite forces.
Formula Used: The formulae used in the solution are given here.
${F_1} = \dfrac{{{\mu _0}{I_1}{I_2}l}}{{2\pi r}}$ where ${\mu _0}$ is the permeability of free space, ${I_1}$ and ${I_2}$ are currents passing through the wires, $l$ is the length of the side of the loop and $r$ is the distance of the sides of the loop.

Complete Step by Step Solution: When the loop is placed with the longest side parallel to the long straight wire, then the shortest sides of the loop at the same distance from the wire and force on them are equal but in opposite directions therefore net force on two short sides are zero.
It has been given that, a rectangular loop of sides 25 cm and 10 cm carrying current of 15 A is placed with its longer side parallel to a long straight conductor $2.0cm$ apart carrying a current of 25 A.
Now longest side which near the wire, force on it is given by ${F_1} = \dfrac{{{\mu _0}{I_1}{I_2}l}}{{2\pi {r_1}}} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 25 \times 15 \times 0.25}}{{2\pi \times 0.02}} = 9.375 \times {10^{ - 4}}N$.
The current in the wire parallel to the longer side and the current in the longer side are in the same direction, so the wire is attracted towards the wire that forms the long side of the loop. This attractive force is given by,
${F_2} = \dfrac{{{\mu _0}{I_1}{I_2}l}}{{2\pi {r_2}}} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 25 \times 15 \times 0.25}}{{2\pi \times 0.12}} = 1.563 \times {10^{ - 4}}N$.
Since, the current flow along the shorter sides of the rectangular loop is equal and opposite, by symmetry they exert equal and opposite forces (${F_3}$ and ${F_4}$). Hence net force on these sides is zero.
Net force is thus given by the difference of forces ${F_1}$ and ${F_2}$. Subtraction is done due to repulsion.
$F = {F_1} - {F_2} = 0$
$ = \left( {9.375 - 1.563} \right) \times {10^{ - 4}} = 7.812 \times {10^{ - 4}}N$.
Thus, the net force is directed away from long wire.

Note: When an electric current passes through a wire, it creates a magnetic field around it. Also, this magnetic field forms concentric circles around the wire. Furthermore, the direction of the magnetic field depends upon the direction of the current.
Moreover, we can determine it by using the ‘right-hand rule’, by pointing the thumb of your right hand in the direction of the current. Besides, the direction of magnetic field lines in the direction of your curled fingers. And the magnitude of field depends on the amount of current, and the distance from the charge-carrying wire.
Fleming's left hand rule is used to determine the direction of force exerted on a current carrying wire placed in a magnetic field. If the thumb, index finger (along magnetic field) and middle finger (along current) are held mutually perpendicular as shown in the figure, then the thumb gives the direction of force on the wire.