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A metallic rod of length 'l' is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present every where. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it. |
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Answer» Solution :We know that a conductor has free charges. If a conductor of LENGTH .I. is moving perpendicular to a uniform magnetic field `vecB` with a velocity `vecv` perpendicular to its length too, then a Lorentz force `vecF` acts on each free charge, where `|vecF| = q vB` Due to this force free charges move along the length of conductor from one end to the other end and exact direction is given by Fleming.s left HAND rule. As a result of this motion of charge, a potential difference is set up between the two ends of conductor. The potential difference developed (i.e., the induced emf) is given by `|varepsilon| = (FI)/q = (qvBl)/q=vBl` Now, consider a conductor rod of length .l. hinged at one end Mand rotating about a NORMAL axis passing through Mwith a frequency v, then for a small element of rod of thickness .de. SITUATED at a distance .x. from the hinged end, the induced emf is `dvarepsilon = Bv DX = B (xomega)dx = B (x.2piv) dx = 2pivB xdx` `therefore` Total induced emf `varepsilon =2pivB int_(0)^(1) xdx = 2pivB. [l^(2)/2] = pivBl^(2)` 
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