Question

A long solenoid with 9.95 turns/cm and a radius of 6.45 cm carries a current of 17.2 mA. A current of 2.55 A exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis in centimeters will the direction of the resulting magnetic field be at 49.4° to the axial direction? (b) What is the magnitude of the magnetic field there?

Answer #1

n = number of turns per unit length = 9.95 x 100 turns/m = 995

i_{s} = current in solenoid = 0.0172 A

magnetic field by the solenoid is given as

B_{s} = u_{o} ni_{s} = (12.56 x
10^{-7}) (995) (0.0172) = 2.15 x 10^{-5}
T parallel to axis

B = resultant magnetic field

B_{w} = magnetic field by the wire

B_{s} = B Cos49.4

2.15 x 10^{-5} = B Cos49.4

B = 3.3 x 10^{-5} T

for the wire

r = distance from the wire

i_{w} = current in wire = 2.55

magnetic field by the wire is given as

B_{w} = (/4)
(2i_{w}/r)

B sin49.4 = (/4)
(2i_{w}/r)

(3.3 x 10^{-5} ) sin49.4 = (10^{-7}) (2
(2.55)/r)

r = 0.0204 m

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