NEW PAGE 1. In Figure 1, the currents in the long, straight wire is I -600 A and

Question

NEW PAGE 1. In Figure 1, the currents in the long, straight wire is I -600 A and the wire lies in the p loop, which carries 15.0 A. The dimension shown are a = 0.300 m, c = 0.120 m, and L a. Find the magnitude and direction of the force exerted by the magnetic field due b. Find the sum of the magnitudes and directions of the forces exerted by the magneti c. Find the magnitude and direction of the force exerted by the magnetic field due to the magnitude and direction of the net force exerted by the magnetic field due to t segment i-ii of the loop. straight wire segments ii-ii and iv-i of the loop segment ii-iv of the loop. d. Find the on the whole loop. Figure 1 Ii

Explanation / Answer

parallel current attracts and anti parallel Currents repel.

part A:

from bottom part of loop, constitutes anti parallel currents with wires being separated by a distance c

assuming L = 1 m, ( as this part is missed in your question)

force between parallel wires is given

F = uo I1 I2 * L/(2pid)

here for part a: d =c

so

F = (4pi *10^-7 *6* 15 *1)/(2 pi* 0.12)

Force due to segment i-ii = 15*10^-5 N,   repulsive force .

part b:

force due to segments ii-iii and iv-i is zero as currents are opposite to each other

F due upward current is cancelled by force due to downward current

part C:

segment iii-iv corresponds to parallel current with distance between the wires = a

so

F = (4pi *10^-7 *6* 15 *1)/(2 pi* 0.3)

F =   6*10^-5 N attractive force

part D:

Net force = F top + Fbottom

now here top loop of wire is attracting and bottom loop of wire is repelling

so
Fnet = 15*10^-5 - 6 *10^-5

Fnet = 9 *10^-5 N    Direction is away from the loop

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