A long solenoid with 10.0 turns/cm and aradius of 7.00 cm carries a current of28

Question

A long solenoid with 10.0 turns/cm and aradius of 7.00 cm carries a current of28.0 mA. A current of 5.00 A exists in a straight conductor located alongthe central axis of the solenoid. (a) At what radial distance from the axis willthe direction of the resulting magnetic field be at 45.0° tothe axial direction?
1 cm

(b) What is the magnitude of the magnetic field there?
2 T (a) At what radial distance from the axis willthe direction of the resulting magnetic field be at 45.0° tothe axial direction?
1 cm

(b) What is the magnitude of the magnetic field there?
2 T

Explanation / Answer

a) The magnetic field due to the very long wire           Bwire = µoIwire/2r. The resulting magnetic field be at 45.0° to the axialdirection. Therefore       Bsolenoid = Bwire µonIsolenoid = µoIwire/2 r ==>   r = (Iwire/Isolenoid)/2n = (5.0/28 x10-3)/(2 *1000 * 3.14 ) = 28.43 x 10-3m            = 2.843 cm b) For a very long solenoid, the magnetic field inside the solenoidis Bsolenoid =µonIsolenoid,                  B = (4 x 10-7 N/A2)(1000/m)(28 x10-3 A)    = 3.516 x 10-5 N/A-m                     = 3.516 x 10-5 T.

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