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A closely wound solenoid of 800 turns and area of cross-section 2.5 xx 10^(-4) m^2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Answer»

Solution :Here N= 800 , A = `2.8xx10^(-4) m^2 ` and I = 3.0 A
`therefore ` MAGNETIC MOMENT of solenoid m = NIA =` 800xx3.0xx2.5xx10^(-4)=0.60 A m^2`
The axis of the solenoid acts like the axis of a bar magnet . One end of solenoid BEHAVES as N-pole and the other end as S-pole . Exact direction of FIELD can be determined by applying the right hand rule.


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