Question

There are 2 wires, i_1 = 3A () and i_2 = 5A (). The wires are 15 cm apart on the y-axis and there is a point P 10 cm to the right of i_1 on the x-axis Find the magnitude & direction of B rightarrow at pt P. (remembeFind theg that B is a vector). We have the same 2 wires viewed looking down on them. A loop with current is to the right as shown. Find the net magnitude force on the loop. You can ignore other forces(like gravity) Think carefully about B rightarrow there are for the loop to interact with. This is NOT the equation for Force between 2 wires alone.

The symple (o) means out of the page

Answer 1

Magnetic field on point P:

$B=\frac{u_{0}}{2\pi }(\frac{I_{1}}{r_{1}}+\frac{I_{2}}{r_{2}})$

$B=\frac{4\pi X10^{-7}}{2\pi }(\frac{3}{0.1}+\frac{5}{\sqrt{0.1^{2}+0.15^{2}}})$

$\boldsymbol{B=11.55X10^{-6}\: T}$

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B)

The force between two perpendicular current carrying wires is zero. Only the sides of the square parallel to the inifinite wire contribute to the force

Force on loop by wire 1

$F_{1}=h*I*B_{1}$

$F_{1}=1.3*100*(\frac{(4\pi X10^{-7})*3}{2\pi *(12.5+0.15)}+\frac{(4\pi X10^{-7})*3}{2\pi *(12.5+0.15+0.5)})$

$F_{1}=12.10X10^{-6}N$

By wire 2

$F_{2}=1.3*100*(\frac{(4\pi X10^{-7})*5}{2\pi *(12.5)}+\frac{(4\pi X10^{-7})*5}{2\pi *(12.5+0.5)})$

$F_{2}=20.4X10^{-6}N$

Total force is the sum of 1 and 2

$\boldsymbol{F_{T}=32.5X10^{-6}N}$