A) \[0.6\times {{10}^{7}}N\]
B) \[0.06\times {{10}^{7}}N\]
C) \[0.6N\]
D) \[0.06N\]
Correct Answer: C
Solution :
\[F=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{6M{M}'}{{{d}^{4}}} \right)\] in end-on position between two small magnets. \[\therefore \ F={{10}^{-7}}\left( \frac{6\times 10\times 10}{{{\left( 0.1 \right)}^{4}}} \right)=0.6N\]
You need to login to perform this action.
You will be redirected in
3 sec
