A) 0.1 T
B) 0.2 T
C) 0.3 T
D) 0.4 T
Correct Answer: A
Solution :
Point P lies on equatorial line of magnet and axial line of magnet as shown \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{M}{{{d}^{3}}}\,=\,\,{{10}^{-7}}\,\times \,\,\frac{1000}{{{(0.1)}^{3}}}\,\,=\,\,0.1\,T\] \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2M}{{{d}^{3}}}\,\,=\,{{10}^{-7}}\,\times \,\,\frac{2\times 1000}{{{(0.1)}^{3}}}\,\,=\,\,0.2\,T\] \[\therefore \,\,\,\,{{B}_{net}}={{B}_{2}}-{{B}_{1}}=0.1\,T\]You need to login to perform this action.
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