| Two parallel vertical metallic rails AB and CD are separated by 1m. They are connected at the two ends by resistances \[{{R}_{1}}\]and \[{{R}_{2}}\] as shown in figure. Horizontal metallic bar of mass \[0.2\,kg\] slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the power dissipated in \[{{R}_{1}}\]and \[{{R}_{2}}\] are \[0.76\,W\] and \[1.2\,W\] respectively |
 |
| \[{{R}_{1}}\] is |
A) \[\frac{9}{19}\Omega \]
B) \[0.2\Omega \]
C) \[5\Omega \]
D) \[0.3\Omega \]
Correct Answer: A
Solution :
| At terminal velocity magnetic force = weight |
| \[BII=mg\,i.e.\,.l=\frac{0.2\times 9.8}{0.6\times 1}=\frac{9.8}{3}A\] |
| Now if e is the emf induced in the rod. |
| \[e\times l=P={{P}_{1}}+{{P}_{2}};e=\frac{(0.76+1.20)}{(9.8/3)}=0.6V\] |
| Further, as in case of Joule heating |
| \[P=\frac{{{V}^{2}}}{R}i.e.,R=\frac{{{V}^{2}}}{P};\] and as here , \[{{V}_{1}}={{V}_{2}}=e\] |
| So, \[{{R}_{1}}=\frac{{{e}^{2}}}{{{p}_{1}}}=\frac{{{(0.6)}^{2}}}{0.76}=\frac{9}{19}\Omega \] |
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