question_answer

           

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,                                     \[{{B}_{net}}=\sqrt{{{\left( {{B}_{H}}-{{B}_{wire}} \right)}^{2}}+B_{V}^{2}}\] \[{{B}_{net}}=\sqrt{{{\left( 0.12 \right)}^{2}}+{{(0.22)}^{2}}}=0.25G\]