A closely wound solenoid of 800 turns and area of cross section `2*5xx10^-4m^2` carries a current of `3*0A`. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Explanation

The magnetic moment of the solenoid of N turns, area of cross-section A and carrying current I is given by:
M = NIA
Here, `N = 800 , A = 2.5 xx 10^(-4) m^2 , I = 3.0 A`
` therefore M = 800 xx 3 xx 2.5 xx 10^(-4) = 0.60 JT^(-1)`
along the axis of the solenoid, determined by the sense of current as per the right-hand screw rule. The solenoid acts like a bar magnet in the sense that just like in a bar magnet the magnetic field lines originate from one pole and goes to the other end. Similarly, in the solenoid, the magnetic field is produced along its axis and field lines originate from one end and go to the other end. So, the two ends of solenoid act as two poles of a bar magnet.