A power line carries a DC current of I = 29 A in a direction = 41° east of magne

Question

A power line carries a DC current of I = 29 A in a direction = 41° east of magnetic north through an open field, in a location where the Earth’s magnetic field is horizontal and its strength is B = 0.61 G.

D: In what direction(s) could you orient the current flow to maximize the magnetic force acting on the wire?

E: In which of the directions that correctly answer part (d) is the magnetic force on the wire up?

F: If the wire is oriented so that the current flows in the direction you chose for part (e), what would the mass per unit length of the wire, , in grams per meter, need to be for the magnetic force to balance the weight of the wire?

Explanation / Answer

magnetic north is along -j.

(D) F = IL X B = I L B sin(theta)

when theta = 90 deg then , force will be maximum.

so when current is to east or west.

(or current is up or down)

(E) F -> k

and B -> -j

then IL^ will be along -i.

hence to the west.

(F) I L B = (mu L ) g

I B = mu g

29 x 0.61 x 10^-4 = mu (9.8)

mu = 1.805 x 10^-4 kg/m

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