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INDIA’S ATOMIC ENERGY PROGRAMME

 The atomic energy programme in India was launched around the time of independence under the leadership of Homi J. Bhabha (1909-1966). An early historic achievement was the design and construction of the first nuclear reactor in India (named Apsara) which went critical on August 4, 1956. It used enriched uranium as fuel and water as moderator. Following this was another notable landmark: the construction of CIRUS (Canada India Research U.S.) reactor in 1960. This 40 MW reactor used natural uranium as fuel and heavy water as moderator. Apsara and CIRUS spurred research in a wide range of areas of basic and applied nuclear science. An important milestone in the first two decades of the programme was the indigenous design and construction of the plutonium plant at Trombay, which ushered in the technology of fuel reprocessing (separating useful fissile and fertile nuclear materials from the spent fuel of a reactor) in India. Research reactors that have been subsequently commissioned in...
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Magnetic force on a current-carrying conductor

 We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA.

For a steady current I in this conducting rod, we may assume that each mobile carrier has an average

drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is:

F = (nlA)q vd × B

where q is the value of the charge on a carrier. Now nq vd is the current density j and |(nq vd )|A is the current I (see Chapter 3 for the discussion of current and current density). Thus,

F = [(nq vd )lA] × B = [ jAl ] × B = Il × B

where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to l. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dl j and summing

F = ∑Id × l B

This summation can be converted to an integral in most cases 

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