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Question: A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in ...

A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The Earth’s magnetic field at the place is 0.39 G and the angle of dip is 35{{35}^{{}^\circ }}. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Explanation

Solution

The horizontal and vertical components of magnetic fields can be calculated using the given values of Earth’s magnetic field at a given point and the angle of dip. Use the formula Bwire=μ04πIr{{B}_{wire}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{r} to find out the magnetic field produced due to the current flowing in a telephone cable. Finally, do a vector sum of all these magnetic fields to get a resultant magnetic field at a given point.

Complete answer:

The figure illustrates the situation mentioned in the question.
Given that,
The current in the wire = I = 1.0 A
Earth’s magnetic field = B = 0.39 G
The angle of deep = δ=35\delta ={{35}^{{}^\circ }}
The horizontal component of Earth’s magnetic field is given as
BH=Bcosδ BH=0.39cos35 BH=0.39×0.8139 BH=0.32G \begin{aligned} & {{B}_{H}}=B\cos \delta \\\ & {{B}_{H}}=0.39\cos {{35}^{{}^\circ }} \\\ & {{B}_{H}}=0.39\times 0.8139 \\\ & \therefore {{B}_{H}}=0.32G \\\ \end{aligned}
The vertical component of Earth’s magnetic field is given as
BV=Bsinδ BV=0.39sin35 BV=0.39×0.5736 BV=0.22G \begin{aligned} & {{B}_{V}}=B\sin \delta \\\ & {{B}_{V}}=0.39\sin {{35}^{{}^\circ }} \\\ & {{B}_{V}}=0.39\times 0.5736 \\\ & \therefore {{B}_{V}}=0.22G \\\ \end{aligned}
The magnetic field due to the current in telephone cable is given as
Bwire=μ04πIr{{B}_{wire}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{r}
We have to find the resultant magnetic fields at points 4.0 cm below the cable i.e. at r = 4.0 cm
Therefore,
Bwire=107×14×102=2×105T=0.2G{{B}_{wire}}={{10}^{-7}}\times \dfrac{1}{4\times {{10}^{-2}}}=2\times {{10}^{-5}}T=0.2G
The net magnetic field at this point is given as
Bnet=(BHBwire)2+BV2 Bnet=(0.320.2)2+0.222 Bnet=0.25G \begin{aligned} & {{B}_{net}}=\sqrt{{{({{B}_{H}}-{{B}_{wire}})}^{2}}+{{B}_{V}}^{2}} \\\ & {{B}_{net}}=\sqrt{{{(0.32-0.2)}^{2}}+{{0.22}^{2}}} \\\ & \therefore {{B}_{net}}=0.25G \\\ \end{aligned}
Hence, the resultant magnetic fields at points 4.0 cm below the cable is 0.25 G.

Note: Choose the units of all the quantities in the same system, later convert into the appropriate form of unit. Gauss (G) is the CGS unit of magnetic field whereas Tesla (T) is the SI unit of the magnetic field.
1T=10-6G\text{1T=1}{{\text{0}}^{\text{-6}}}\text{G}. The direction of the magnetic field produced by a current flowing through a straight wire is given by right hand thumb rule.