A telephone cable at a place has four long straight horizontal wires carrying a current of 4.0 A in the same direction east to west. |
The earths magnetic field at the place is 0.39 G, and the angle of dip is \[35{}^\circ \]. The magnetic declination is nearly zero. What is the resultant magnetic field at points 4.0 cm below the cable? |
A) 0.25 G
B) 2.32 G
C) 1.93 G
D) 3.11G
Correct Answer: A
Solution :
[a] : Let us first decide the directions which can best represent Qwest the situation, Here, \[{{B}_{H}}=B\cos \delta \] \[=0.39\times \cos {{35}^{o}}\] \[{{B}_{H}}=0.32G\]and \[{{B}_{V}}=B\sin \delta \] \[=0.39\times \sin 35{}^\circ \] \[{{B}_{V}}=0.22\text{ }G\] Telephone cable carry a total current of 4.0 A in direction east to west. We want resultant magnetic field 4.0 cm below. Now,\[{{B}_{Wire}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2I}{r}\] \[={{10}^{-7}}\times \frac{2\times 4}{4\times {{10}^{-2}}}\] \[=2\times {{10}^{-5}}T=0.2G\] Net magnetic field, \[{{B}_{net}}=\sqrt{{{\left( {{B}_{H}}-{{B}_{wire}} \right)}^{2}}+B_{V}^{2}}\] \[{{B}_{net}}=\sqrt{{{\left( 0.12 \right)}^{2}}+{{(0.22)}^{2}}}=0.25G\]You need to login to perform this action.
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