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A uniform magnetic fieldvecB is set up along the positive x-axis. A particle of charge 'q' and mass 'm' moving with a velocity vecv entresthe field at the origin in X-Y plane such that it has velocity companents both along and perpendicular to the magnetic field vecB. Trace, giving reason. the fraectory followed by the particle . Find out the expression for the distance moved by the particle along the magnetic field in one rotation. |
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Answer» Solution :field at an axial point OFA circular coil: ` B= oint DB sin theta` where, `dB= mu_(0)/(4PI) (idl)/x^(2)` Where N is the number of TURNS of the circular coil. If M -iA is the magnetic moment of the coil, then ` sin theta = a/x Rightarrow B = mu_(0)/(4pi) (ia)/x^(3) oint dl = mu_(0)/(4pi) ( ia (2pia))/ ( r^(2) +a^(2))^(3//2)` ` B= (mu_(0) ia^(2) N)/(2 (r^(2) +a^(2))^(3//2))` Where, N is the number of turns of the circular, coil. If M =iA is the magnetic moment of thecoil, then `M = i pia^(2) ` for single turn ` M= pia^(2)` for a coil of N turns ` B = mu_(0)/(4pi) .(2M)/((r^(2) +a^(2))^(3//2)) ` for small loop, ` B =(mu_(0))/(4pi) .(2M)/r^(3) Wb // m^(2)` This is a SIMILAR result as obtained end on position of electric dipole where, ` E= 1/ (4pie_(0)). (2p)/(r^(3)) ` ( where is electric moment. 
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