Shown above is the cross section of a very long solenoid with N=5,190 turns. hav

Question

Shown above is the cross section of a very long solenoid with N=5,190 turns. having length L=0.560 m, radius R=0.0800 m and carrying positive current i>0. The B field inside the solenoid due to the solenoid current i alone has the magnitude B=0.000590 T and is parallel to the axis of the solenoid. Also shown is a long straight wire running along the axis of the solenoid carrying current I=45.9 A into the page.

a) What is the magnitude of the total B field (due to both solenoid and wire) at a distance r=0.0250 m from the wire?

Explanation / Answer

Given: No. of turns of solenoid = N = 51900 turns Lenght of the solenoid = L = 0.560m Radious of the solenoid = R = 0.08 m Magnetic field inside the solenoid = BS = 0.000590 T current carrying in the wire = I = 45.9 A distance from the centrer of the solenoid to wire = r = 0.0250 m    (a) magnitude of the total B field (due to both solenoid and wire)   at a distance r=0.0250 m from the wire is B = BS+ Bw      where Bs magnetic field due to solenoid ,              Bw magnetic field due to wire      = (Bs) + o I / 2 r      = ( 0.000590 ) + [(4x 10-7 x 45.9/ 2(0.025) ]      =  0.000590 T + 0.0003672 T      = 0.0009572 T             or
B = 9.572 x 10-4 T

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