A) \[4\times {{10}^{-5}}wb/{{m}^{2}}\]
B) \[2\times {{10}^{-5}}wb/{{m}^{2}}\]
C) \[3\times {{10}^{-5}}wb/{{m}^{2}}\]
D) \[8\times {{10}^{-5}}wb/{{m}^{2}}\]
Correct Answer: A
Solution :
Here: \[~i=20\] amp, \[r=10\text{ }cm=10\times {{10}^{-2}}m\]Intensity of magnetic field produced due to straight current carrying wire will be \[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{r}\] \[=\frac{{{10}^{-7}}\times 2\times 20}{0.1}=4\times {{10}^{-5}}\,wb/{{m}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
