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(a) Derive the expression for the torque acting on the rectangular current carrying coll of a galvanometer. Why is the magnetic field made radial ? (b) A alpha- particle is accelerated through a potential difference of 10 kV and moves along y-axis. It enters in a region of uniform magnetic field B=2 xx 10^(-3)T acting along y-axis. Find the radius of its path. ["Take mass of "alpha-"particle "=6.4 xx 10^(-27)kg] |
Answer» Solution :(a) Consider a rectangular galvanometer coil PQRS of length land breadth b, carrying a CURRENT L, 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 dB.  In this situation forces F and E, having magnitude Ib 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 ll B 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. Figure (b) represents a top view of the arrangement and from it, we find that the torque of the couple acting on the loop. `tau=(II B) xx b sin theta=l(l b ) B sin theta=IA B sin theta` where A =l b= area of the loop. If anstead of a single loop, we have a rectangular coil having N turns, then torque `tau=NIA B sin theta` In vector notation `vectau=NI( VECA xx vecB)=(vecm xx vecB)," where "vecm=NI vecA`= magnetic moment of current carrying coil. The magnetic field is made radial so that value of `theta` BECOMES `90^@` and in that case torque `tau=NIAB`, whose value does not change even when the coil rotates by a certain angle. (b) Here accelerating voltage `V =10kV =10 xx 10^(3) V,` normal magnetic field `B= 2 xx 10^(-3)T` mass of `alpha` -particles `m=6.4 xx 10^(-27)kg` and charge of `alpha`-particle `q=+2e=2 xx 1.6 xx 10^(-19) =3.2 x 10^(-19)C` Radius of circular path `r=sqrt((2mV)/(qB^(2))) =sqrt(2 xx (6.4 xx 10^(-27)) xx (10 xx 10^(3))/((3.2 xx 10^(-19)) xx (2 xx 10^(-3))^(2)))=10m` |
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