A) \[2.23\times {{10}^{-2}}\]
B) \[2.23\times {{10}^{-3}}\]
C) \[2.25\times {{10}^{-3}}\]
D) \[2.26\times {{10}^{-2}}\]
Correct Answer: B
Solution :
Two identical short bar magnets are perpendicular to each other and centres are at a distance 0.2 m from each other. Magnetic moment of each magnet \[=M=10A{{m}^{2}}\] Magnetic induction at point P due to first magnet is \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{(0.1)}^{3}}}\] (end on position) Similarly magnetic induction at point P due to second magnet is \[{{B}_{2}}=\frac{{{\mu }_{_{0}}}}{4\pi }\frac{M}{{{(0.1)}^{3}}}\](broad side on position) Since, both \[{{B}_{1}}\]and \[{{B}_{2}}\]are mutually perpendicular to each other, so the resultant magnetic induction at point P is \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=\sqrt{\left[ \frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{(0.1)}^{3}}} \right]+{{\left[ \frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{(0.1)}^{3}}} \right]}^{2}}}\] \[B=\sqrt{{{\left[ \frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{(0.1)}^{3}}} \right]}^{2}}[{{2}^{2}}+1]}\] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{(0.1)}^{3}}}\sqrt{4+1}\] \[={{10}^{-7}}\times M\times {{10}^{3}}\sqrt{5}\] Here, \[M=10A{{m}^{2}}\] \[B={{10}^{-7}}\times 10\times {{10}^{3}}\times 2.23\] \[B=2.23\times {{10}^{-3}}T\]
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