Question #175128

A long solenoid of cross-sectional area 28๐‘๐‘š2

A coil ๐ถ of 160 turns of insulated wire is wound tightly around the centre of the solenoid. If ๐‘› is a 

constant equal to 1.5 ร— 10

3๐‘šโˆ’1

and ๐œ‡๐‘œ is the permeability of free space, calculate for a current of 

3.5๐ด in the solenoid; (i) the magnetic flux at the centre of the solenoid. (ii) the flux linkage in the 

coil ๐ถ. (iii) If the current in the coil ๐ถ is reversed in 0.80๐‘ , calculate the average e.m.f. induced in 

the coil ๐ถ.


1
Expert's answer
2021-03-29T09:03:51-0400

(i) The magnetic flux at the centre of the solenoid:


B=ฮผ0nI=4ฯ€โ‹…10โˆ’7โ‹…1.5โ‹…103โ‹…3.5=6.6โ‹…10โˆ’3 T.B=\mu_0nI=4\piยท10^{-7}ยท1.5ยท10^3ยท3.5=6.6ยท10^{-3}\text{ T}.

(ii) The flux linkage in the coil ๐ถ:


ฮฆ=BAN=6.6โ‹…10โˆ’3โ‹…25โ‹…10โˆ’4โ‹…160==3โ‹…10โˆ’3 Wb.\Phi=BAN=6.6ยท10^{-3}ยท25ยท10^{-4}ยท160=\\=3ยท10^{-3}\text{ Wb}.


(iii) If the current in the coil ๐ถ is reversed in 0.80๐‘ , calculate the average e.m.f. induced in the coil ๐ถ:


e.m.f.=ฮ”ฮฆฮ”t=2ฮฆฮ”t=7.5โ‹…10โˆ’3 V.\text{e.m.f.}=\frac{\Delta \Phi}{\Delta t}=\frac{2\Phi}{\Delta t}=7.5ยท10^{-3}\text{ V}.



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