Question #175128

A long solenoid of cross-sectional area 28π‘π‘š2

A coil 𝐢 of 160 turns of insulated wire is wound tightly around the centre of the solenoid. If 𝑛 is a 

constant equal to 1.5 Γ— 10

3π‘šβˆ’1

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

3.5𝐴 in the solenoid; (i) the magnetic flux at the centre of the solenoid. (ii) the flux linkage in the 

coil 𝐢. (iii) If the current in the coil 𝐢 is reversed in 0.80𝑠, calculate the average e.m.f. induced in 

the coil 𝐢.


Expert's answer

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


B=ΞΌ0nI=4Ο€β‹…10βˆ’7β‹…1.5β‹…103β‹…3.5=6.6β‹…10βˆ’3 T.B=\mu_0nI=4\piΒ·10^{-7}Β·1.5Β·10^3Β·3.5=6.6Β·10^{-3}\text{ T}.

(ii) The flux linkage in the coil 𝐢:


Ξ¦=BAN=6.6β‹…10βˆ’3β‹…25β‹…10βˆ’4β‹…160==3β‹…10βˆ’3 Wb.\Phi=BAN=6.6Β·10^{-3}Β·25Β·10^{-4}Β·160=\\=3Β·10^{-3}\text{ Wb}.


(iii) If the current in the coil 𝐢 is reversed in 0.80𝑠, calculate the average e.m.f. induced in the coil 𝐢:


e.m.f.=ΔΦΔt=2ΦΔt=7.5β‹…10βˆ’3 V.\text{e.m.f.}=\frac{\Delta \Phi}{\Delta t}=\frac{2\Phi}{\Delta t}=7.5Β·10^{-3}\text{ V}.



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