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Nookie1986 [14]
3 years ago
11

A long wire is known to have a radius greater than 4.0 mm and to carry a current uniformly distributed over its cross section. i

f the magnitude of the magnetic field is 0.285 mt at a point 4.0 mm from the axis of the wire and 0.200 mt at a point 10 mm from the axis, what is the radius of the wire?
Physics
1 answer:
ollegr [7]3 years ago
8 0

Magnetic field outside it due to long wire is given by

B = \frac{u_o i}{2 \pi r}

Magnetic field due to long wire inside wire at any point

B = \frac{u_o i r}{2 \pi R^2}

Now the ratio of two magnetic field is given by

\frac{B_{in}}{B_{out}} = \frac{r_1/R^2}{1/r_2}

\frac{0.285}{0.200} = \frac{4*10}{R^2}

1.425 = \frac{40}{R^2}

R = 5.3 mm

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<u>x-values:</u>

<u />x_{cm} = \frac{m_{1}x_{1}   + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}} = \frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12} = \frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41} = \frac{81}{41} = 1.9756

<u>y-values:</u>

<u />y_{cm} = \frac{m_{1}y_{1}   + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}} = \frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12} = \frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41} = \frac{-90}{41} = -2.1951

<u>center of mass:</u>

(1.9756, -2.1951)

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