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A solenoid 60 cm long and of radius 4.0 cm has a layers of windings of 300 turns each. A 2.0cm long wire of mass 2.5 g lies inside the solenoid (near itscentre) normal to its axis, both the wire and the axis of the solenoid are in the horizontal plane. The wireis connected through two leads parallel to the axis of hte solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (With appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? g = 9.8 m s^(-2). |
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Answer» Solution :Here `l = 60 cm= 0.6 m, ` total number of turns in SOLENOID coil `N = 3 xx 300 = 900`, HENCE number of turns per unit length `n = N/l = 900/0.6 = 1500 m^(-1)` , current FLOWING in coil = I, mass of wire `m = 2.5 g= 2.5 xx 10^(-3) kg`, current flowing in wire `I. = 6 A` and length of wire `l. = 2 cm = 2 xx 10^(-2) m` `:.` Magnetic field inside the solenoid near its CENTRE `B = mu_0 n I = 4 pi xx 10^(-7) xx 1500 xx I = 6 pi I xx 10^(-4) T` The magnetic field `vecB` acts along the axis of solenoid and is, thus, normal to the wire. Therefore, the force experienced by the wire is `F = B I. l. = (6 pi xx 10^(-4) I) xx (6) xx (2 xx 10^(-2)) = 72pi xx 10^(-6) I N` This force should just support the weight of the wire `:. 72 pi xx 10^(-6) I = (2.5 xx 10^(-3)) xx (9.8)` `I = (2.5 xx 10^(-3) xx 9.8)/(72 pi xx 10^(-6)) = 108.37 A ~~ 108 A` |
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