Question

In Figure 20.15, a rectangular current loop is carrying current I1 = 7.0 A, in the direction indicated near a long wire carrying a current Iw. The long wire is parallel to the sides of the rectangle. The rectangle loop has length 0.80 m and its sides are 0.10 m and 0.70 m from the wire. If the net force on the loop is to have magnitude 1.7x10^-6 and is to be directed towards the wire, what must be the (a) magnitude and (b) direction (from top to bottom or from bottom to top in the sketch) of the current Iw in the wire?

Note that:

\(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} .\)

Answer 1

current passing through the loop,

$$ I_{1}=7.0 \mathrm{~A} $$

current passing through the long wire,

$$ I_{w}=? $$

lenght of the loop,

$$ L=0.8 \mathrm{~m} $$

distance between wire to first vertical side of the loop,

$$ p=0.1 \mathrm{~m} $$

distance between two vertical sides of the loop is,

$$ \begin{aligned} q &=0.7 \mathrm{~m}-0.1 \mathrm{~m} \\ &=0.6 \mathrm{~m} \end{aligned} $$

Net force due to wire on the loop is,

$$ F=1.7 \times 10^{-6} \mathrm{~N} $$

Mangetic force between wire to first vertical side of the loop is,

$$ F_{1}=\frac{\mu_{o} I_{1} I_{w} L}{2 \pi p} $$

Mangetic force between wire to second vertical side

of the loop is, \(F_{2}=\frac{\mu_{0}\left(-I_{1}\right) I_{w} L}{2 \pi(p+q)} \quad(-\) sign is for current direction \()\)

Here, it should be note that, the magnetic force due to wire and paralle sides of the loop contributes 0 as the current in them is opposite to each other and perpenducular to wire's current direction

Thus, resutant force between wire and loop is,

$$ \begin{aligned} F &=F_{1}+F_{2} \\ &=\frac{\mu_{0} l_{1} I_{w} L}{2 \pi p}+\frac{\mu_{0}\left(-I_{1}\right) I_{w} L}{2 \pi(p+q)} \\ &=\frac{\mu_{0} I_{1} l_{w} L}{2 \pi}\left(\frac{1}{p}-\frac{1}{p+q}\right) \\ &=\frac{\mu_{0} l_{1} l_{w} L}{2 \pi}\left[\frac{q}{p(p+q)}\right] \\ =& \frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(7.0 \mathrm{~A}) I_{\mathrm{w}}(0.80 \mathrm{~m})}{2 \pi}\left[\frac{(0.6 \mathrm{~m})}{(0.1 \mathrm{~m})(0.1 \mathrm{~m}+0.6 \mathrm{~m})}\right] \\ =96 \times 10^{-7}\left(I_{\mathrm{w}}\right) \end{aligned} $$

but,

\(F=1.7 \times 10^{-6} \mathrm{~N}\)

Hence, \(\begin{aligned} 96 \times 10^{-7}\left(l_{w}\right) &=1.7 \times 10^{-6} \mathrm{~N} \\ I_{w} &=0.177 \mathrm{~A} \end{aligned}\)

direction of the current is being

bottom to top of the wire