The figure shows, in cross section, four thin wires that are parallel, straight,

Question

The figure shows, in cross section, four thin wires that are parallel, straight, and very long. They carry identical currents of i = 3.25 A in the directions indicated. Initially all four wires are at distance d = 11.9 cm from the origin of the coordinate system, where they create a net magnetic field ModifyingAbove Upper B With right-arrow. (a) To what value of x must you move wire 1 along the x axis in order to rotate ModifyingAbove Upper B With right-arrow counterclockwise by 30°? (b) With wire 1 in that new position, to what value of x must you move wire 3 along the x axis to rotate B by 30° back to its initial orientation?

Explanation / Answer

current passing in the each wire, I=3.25 A

four wires are at a distance(as shown in fig:), d=11.9 cm

=11.9 cm(10^-2 m/1cm)

                                                                            =0.119 m

                         Rotate B counterclockwise by, = 30°

   Magnitude of the magnetic field at the origin due to each current

   carrying wire is,  B = 0I/(2d)
  but,due to current 2 and 4, fields due to 1 and 3 cancel each

other in +x direction.

Hence,Net magentic field is given by,
             Bx = 2B =2[0I/(2d)]       .......(1)

In order to move wire 1 to new position which is distance x

from the origin,

The net field along y direction is,

            By =0I/(2x) - 0I/(2d)      ......(2)

rotation of B in counter clockwise

        tan = By/Bx
from equation(1) , (2)
        tan = (1/x - 1/d)/(2/d)
1/x - 1/d = 2tan/d

             x = d/(1 + 2tan)

put values

                = (0.119 cm)/(1+2tan30o)

                 = 0.0552 m
thus, the new position is,

               x = -0.0552 m

                 (or)

                x=-0.0552 m(10^2 cm/1 m)
                  = -5.52 cm

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