InterviewSolution
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InterviewSolution
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Wire 1 in Fig. is oriented along the y-axis and carries a steady current `I_1`. A rectangular loop located to right of the wire and in the x-y plane carries a current `I_1`. Find the magnetic force exerted by wire (1) on the top wire (1) on the top wire of length b in the loop, labeled "wire (2)" in the figure. |
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Answer» Consider the force exerted by wire 1 on a small segment ds of wire (2). This force is given by `dvecF=IdvecsxxvecB`, where `I=I_2` `vecB` is the magnetic filed created by the current in wire (1) at the position of `dvecs`. The field at a distance x from wire (1) is `B=(mu_0I_1)/(2pix)(-hatk)` where the unit vector `-hatk` is used to idicate that the field due to the current in wire (1) at the position of `dvecs` points into the page. Because wire (2) is along the x-axis, `dvecs=dxhati`, we find that `dF_B=(mu_0I_1I_2)/(2pix)[hatixx(-hatk)]dx= (mu_0I_1I_2)/(2pi) (dx)/x hatj` Integrating over the limits `x=a` to `x=a+b` gives `F_B=(mu_0I_1I_2)/(2pi) lnx]_a^(a+b) hatj` `(mu_0I_1I_2)/(2pi) ln(1+b/a)hatj` The force on wire (2) points in the positive y-direction, as indi- cated by the unit vector `hatj` and as shown in the figure. |
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