A) \[1.5\times {{10}^{-6}}Nm\]
B) \[1.5\times {{10}^{-8}}Nm\]
C) \[1.5\times {{10}^{6}}Nm\]
D) \[1.5\times {{10}^{8}}Nm\]
Correct Answer: B
Solution :
(b) \[1.5\times {{10}^{-8}}N\,m\] Here, for small circular coil, Number of turns, \[N=10\], Area, \[A=1\,m{{m}^{2}}=1\times {{10}^{-6}}{{m}^{2}}\] Current, \[{{l}_{1}}=\frac{21}{44}A\] For a long solenoid, Number of turns per metre, \[n={{10}^{3}}/m\] Current. \[{{l}_{2}}=2.5\,A\] Magnetic field due to a long solenoid on its axis is \[B={{\mu }_{0}}n{{l}_{2}}\] ...(1) Magnetic moment of a circular coil is \[M=NA{{l}_{1}}\] ...(2) Torque, \[\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}\] \[\tau =MB\,\sin \theta =MB[\because \,\theta =90{}^\circ (Given)]\] \[\tau (NA{{l}_{1}})({{\mu }_{0}}n{{l}_{2}})\] [Using eqs. (1) and (2)] \[\tau =10\times 1\times {{10}^{-6}}\times \frac{21}{44}\times 4\times \frac{22}{7}\times {{10}^{-7}}\times {{10}^{3}}\times 2.5\] \[=1.5\times {{10}^{-8}}Nm\]
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