CBSE-XI-Physics
35: Magnetic Field due to a Current
- #9Two particles X and Y having equal charge, after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is
(a) (R1/R2)1/2
(b) R1/R2
(c) (R1/R2)2
(d) R1R2.digAnsr: cAns : (c)`` \frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}``
Particles X and Y of respective masses m1 and m2 are carrying charge q describing circular paths with respective radii R1 and R2 such that
`` {R}_{1}=\frac{{m}_{1}{v}_{1}}{qB}``
`` {R}_{2}=\frac{{m}_{2}{v}_{2}}{qB}``
Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.
`` \therefore \frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{2}{{v}_{2}}^{2}``
`` \because {R}_{1}=\frac{{m}_{1}{v}_{1}}{qB}\Rightarrow {v}_{1}=\frac{{R}_{1}qB}{{m}_{1}}``
`` \,\mathrm{\,And\,},``
`` {R}_{2}=\frac{{m}_{2}{v}_{2}}{qB}\Rightarrow {v}_{2}=\frac{{R}_{2}qB}{{m}_{2}}``
`` \therefore {m}_{1}{\left(\frac{{R}_{1}qB}{{m}_{1}}\right)}^{2}={m}_{2}{\left(\frac{{R}_{2}qB}{{m}_{2}}\right)}^{2}``
`` \Rightarrow \frac{{m}_{1}}{{m}_{2}}=\frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}``
`` ``
Page No 249: