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Obtain the expression for the deflecting torque acting on the current carrying rectangular coil of a galvanometer in a uniform magnetic field. Why is a radial magnetic field employed in the moving coil galvanometer? |
Answer» Solution :Consider a rectangular coil PQRS, of length `l` and breadth b, of a galvanometer carrying a current I, placed in a uniform magnetic field B such that a vector normal to the plane of the coil subtends an angle `theta` from the direction of B.  In this situation forces `F_(1) and F_(2)` having magnitude I b B `sin theta` are acting on arms PQ and RS. The forces are equal, opposite and collinear, hence they cancel out. Again forces `F_(3) and F_(4)` having magnitude `IlB` are acting on each of the two arms QR and SP. These forces too are equal and opposite but these are non-collinear and FORM a couple whose torque is given as: `tau=(IlB)xxb sin theta=I(lb)B sin theta=IAB sin theta` where `A=lb=` area of the LOOP. If instead of a single loop, we have a rectangular coil having N TURNS, then torque `tau=NIAB sin theta` In vector notation `vectau=NI(vecA xx vecB)=(vecm xx vecB)`, where `vecm=NI vecA=` magnetic moment of current carrying coil. A radial magnetic field, in which `theta=90^(@)`, is employed in the moving coil galvanometer because now torque acting on the coil does not change with orientation of the coil. |
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