A telephonic cable at a place has four long straight horizontal wires carrying a current of `1.0amp.` in the same direction east to west. The earth's magnetic field at the place is `0.39G` and the angle of dip is `35^@`. The magnetic declination is almost zero. What are the resultant magnetic fields at points `4.0cm` below and above the cable?

Options

A. 0.25G, 0.55G
B. 0.50G,0.75G
C. 1.25G,0.95G
D. 2.50G,0.45G

Answer

A

Explanation

Let us first decide the direction which can best the direction which can best represent the situation.

`"Here", B_(H)=B cos delta`
`=0.39xx cos 35^(@)G`
`B_(H)=0.32G`
`and B_(V)=B sin delta=0.39xxsin 35^(@)G`
`B_(V)=0.22G`
Telephone cable carry a total curren tof 4.0A in direction east to west. We wnat resultante magnetic field of 4.0cm below.
`B_("wire")=(mu_(0))/(4pi)(2I)/(r)=10^(-7)xx(2xx4)/(4xx10^(-2))`
`=2xx10^(-5)T=0.2G=0.2G`
Net magnetic field
`B_("net")=sqrt((B_(H)-B_("wire")+B_(V)^(2))`
`B_("net")=sqrt((0.12)^(2)+(0.22)^(2))=sqrt(0.0144+0.0484)=0.25G`