A) 3W
B) \[\frac{16}{3R}\]W
C) \[\frac{16}{5R}\]
D) \[\frac{16}{7R}\]
Correct Answer: B
Solution :
The instantaneous moment of the deflecting couple or torque acting on the needle is \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] perpendicular distance = work done When axis of needle makes an angle \[\theta \]with the magnetic field, then for magnetic moment M and magnetic field B, we have \[{{M}_{e}}={{M}_{p}},{{R}_{p}}=\frac{{{R}_{e}}}{4}\] ?...(i) \[\therefore \] ?...(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\frac{{{M}_{e}}}{{{M}_{e}}}\times \frac{{{R}_{e}}}{{{R}_{e}}/4}}=\sqrt{4}=2\] Given, \[\Rightarrow \] \[{{v}_{p}}=2{{v}_{e}}=2\times 11.2\] \[=22.4km/s\]
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