com www.GOALias.blogspot.com

www.GOALias.blogspot.comPhysicsROGET’S SPIRAL FOR ATTRACTION BETWEEN PARALLEL CURRENTSMagnetic effects are generally smaller than electric effects. As a consequence, the forcebetween currents is rather small, because of the smallness of the factor µ. Hence it isdifficult to demonstrate attraction or repulsion between currents. Thus for 5 A currentin each wire at a separation of 1cm, the force per metre would be 5 × 10 –4 N, which isabout 50 mg weight. It would be like pulling a wire by a string going over a pulley towhich a 50 mg weight is attached. The displacement of the wire would be quiteunnoticeable.With the use of a soft spring, we can increase the effective length of the parallel currentand by using mercury, we can make the displacement of even a few mm observable verydramatically. You will also need a constant-currentsupply giving a constant current of about 5 A.Take a soft spring whose natural period ofoscillations is about 0.5 – 1s. Hang it vertically andattach a pointed tip to its lower end, as shown in thefigure here. Take some mercury in a dish and adjust thespring such that the tip is just above the mercurysurface. Take the DC current source, connect one of itsterminals to the upper end of the spring, and dip theother terminal in mercury. If the tip of the spring touchesmercury, the circuit is completed through mercury.Let the DC source be put off to begin with. Let the tip be adjusted so that it justtouches the mercury surface. Switch on the constant current supply, and watch thefascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (justby a mm or so), the circuit is broken, the current stops, the spring relaxes and tries tocome back to its original position, the tip again touches mercury establishing a currentin the circuit, and the cycle continues with tick, tick, tick, . . . . In the beginning, youmay require some small adjustments to get a good effect.Keep your face away from mercury vapours as they are poisonous. Do not inhalemercury vapours for long.Example 4.10 The horizontal component of the earth’s magnetic fieldat a certain place is 3.0 ×10 –5 T and the direction of the field is fromthe geographic south to the geographic north. A very long straightconductor is carrying a steady current of 1A. What is the force perunit length on it when it is placed on a horizontal table and thedirection of the current is (a) east to west; (b) south to north?156EXAMPLE 4.10Solution F = Il × BF = IlB sinθThe force per unit length isf = F/l = I B sinθ(a) When the current is flowing from east to west,θ = 90°Hence,f = I B= 1 × 3 × 10 –5 = 3 × 10 –5 N m –1