Question

A square frame carrying a current $i = 0.90 A$ is located in the same plane as a long straight wire carrying a current $i_{0} = 5.0 A$ . The frame side has a length $a = 8.0 cm$ . The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is $η = 1.5$ times greater than the side of the frame. Find,

$(a)$ ampere force acting on the frame,
$(b)$ the mechanical work to be performed in order to turn the frame through $180 °$ about its axis, with the currents maintained constant.

(Permeability of vacuum,

μ_{0} = 1.257 \times 10^{- 6} H / m

)

Accepted Answer

Given:

A square frame of side, $a = 8.0 cm$ ,

carrying current, $i = 0.9 A$ and

a long straight wire carrying current, $i_{0} = 5.0 A$ .

Solution for $(a)$ :

As it is clear from the condition, Ampere's forces on the sides $(2)$ and $(4)$ are equal in magnitude but opposite in direction.

Hence, the net effective force on the frame is the resultant of the forces, experienced by the sides $(1)$ and $(3)$ .

Now, the Ampere force on $(1)$ ,

$\Rightarrow F_{1} = \frac{μ_{0}}{2 π} \frac{i i_{0}}{(η - \frac{1}{2}) a}$

and that on $(3)$ ,

$\Rightarrow F_{3} = \frac{μ_{0}}{2 π} \frac{i i_{0}}{(η + \frac{1}{2}) a}$

So, the resultant force on the frame,

$\Rightarrow F_{R} = F_{1} - F_{3}$ .
since they are opposite in nature.

$\Rightarrow F_{R} = \frac{2 μ_{0} i i_{0}}{π (4 η^{2} - 1) a}$

$\Rightarrow F_{R} = 0.40 μN$ .

Solution for $(b)$ :

Work done in turning the frame through some angle,

$\Rightarrow A = \int i d Φ = i (Φ_{f} - Φ_{i})$ ,

where

$Φ_{f}$ is the flux through the frame in the final position, and $Φ_{i}$ that in the initial position.

Here, $|Φ_{f}| = |Φ_{i}| = Φ$ and $Φ_{i} = - Φ_{f}$

So, $Δ Φ = 2 Φ$ and

$A = 2 i Φ$

Hence,

$A = 2 i \int \vec{B} \cdot d \vec{S}$

$A = 2 i \int_{a (η - \frac{1}{2})}^{a (η + \frac{1}{2})} \frac{μ_{0}}{2 π} \frac{i_{0} a}{r} d r$

$A = \frac{μ_{0} i i_{0} a}{π} \ln (\frac{2 η + 1}{2 η - 1})$

$∴ A = 0.1 μJ$ .

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