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A wire carrying a current I is bent into the shape of an exponential spiral, r=e^(theta), from theta=0 to theta=2pi as shown in figure (a0 . To complete a loop, the ends of the spiral are connected by a straight wire along the x- axis. Find the magnitude and direction of vec(B) at the origin. |
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Answer» `(mu_(0)I)/(4pi)(1-e^(-2pi))` `tan beta=(r)/((dr)/(d theta))`  Thus in this CASE, we have `r=e^(theta)` and so we get `tan beta =1` and `beta=(pi)/(4)`. Therefore, the angle between `DVEC(1)` and `hat(r)` is `(pi-beta) =(3pi)/(4)` . ALSO`dvec(1)=(dr)/(sin((pi)/(4)))=sqrt(2)dr` From Biot `-` Savart's law, we know that there is no contribution from the straight portion of the wire since `dvec(1)xxvec(r)=0`.For the field of the spiral, we have `dB=(mu_(0)I)/(4pi)((dvec(1)xxhat(r)))/(r^(2))impliesB=(mu_(0)I)/(4pi)int_(theta=0)^(2pi)(|dvec(1)|sin theta|vec(r)|)/(r^(2))` `=(mu_(0)I)/(4pi)int_(theta=0)^(2pi)sqrt(2)dr[sin((3pi)/(4))](1)/(r^(2))` `implies B=(mu_(0)I)/(4pi)int_(theta=0)^(2pi)r^(-2)dr=-(mu_(0)I)/(4pi)(r^(-1))|_(theta=0)^(2pi)` Substitute `r=e^(theta)`, we get `impliesB=-(mu_(0)I)/(4pi)(e^(-theta))|_(0)^(2pi)=-(mu_(0)I)/(4pi)(e^(-2pi)-e^(theta))` `=(mu_(0)I)/(4pi)(1-e^(-2pi))` out of the page. |
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