Answer
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Hint: 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 ${{B}_{wire}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{r}$ 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 = $\delta ={{35}^{{}^\circ }}$
The horizontal component of Earth’s magnetic field is given as
$\begin{align}
& {{B}_{H}}=B\cos \delta \\
& {{B}_{H}}=0.39\cos {{35}^{{}^\circ }} \\
& {{B}_{H}}=0.39\times 0.8139 \\
& \therefore {{B}_{H}}=0.32G \\
\end{align}$
The vertical component of Earth’s magnetic field is given as
$\begin{align}
& {{B}_{V}}=B\sin \delta \\
& {{B}_{V}}=0.39\sin {{35}^{{}^\circ }} \\
& {{B}_{V}}=0.39\times 0.5736 \\
& \therefore {{B}_{V}}=0.22G \\
\end{align}$
The magnetic field due to the current in telephone cable is given as
${{B}_{wire}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{r}$
We have to find the resultant magnetic fields at points 4.0 cm below the cable i.e. at r = 4.0 cm
Therefore,
${{B}_{wire}}={{10}^{-7}}\times \dfrac{1}{4\times {{10}^{-2}}}=2\times {{10}^{-5}}T=0.2G$
The net magnetic field at this point is given as
$\begin{align}
& {{B}_{net}}=\sqrt{{{({{B}_{H}}-{{B}_{wire}})}^{2}}+{{B}_{V}}^{2}} \\
& {{B}_{net}}=\sqrt{{{(0.32-0.2)}^{2}}+{{0.22}^{2}}} \\
& \therefore {{B}_{net}}=0.25G \\
\end{align}$
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.
$\text{1T=1}{{\text{0}}^{\text{-6}}}\text{G}$. The direction of the magnetic field produced by a current flowing through a straight wire is given by right hand thumb rule.
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 = $\delta ={{35}^{{}^\circ }}$
The horizontal component of Earth’s magnetic field is given as
$\begin{align}
& {{B}_{H}}=B\cos \delta \\
& {{B}_{H}}=0.39\cos {{35}^{{}^\circ }} \\
& {{B}_{H}}=0.39\times 0.8139 \\
& \therefore {{B}_{H}}=0.32G \\
\end{align}$
The vertical component of Earth’s magnetic field is given as
$\begin{align}
& {{B}_{V}}=B\sin \delta \\
& {{B}_{V}}=0.39\sin {{35}^{{}^\circ }} \\
& {{B}_{V}}=0.39\times 0.5736 \\
& \therefore {{B}_{V}}=0.22G \\
\end{align}$
The magnetic field due to the current in telephone cable is given as
${{B}_{wire}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{I}{r}$
We have to find the resultant magnetic fields at points 4.0 cm below the cable i.e. at r = 4.0 cm
Therefore,
${{B}_{wire}}={{10}^{-7}}\times \dfrac{1}{4\times {{10}^{-2}}}=2\times {{10}^{-5}}T=0.2G$
The net magnetic field at this point is given as
$\begin{align}
& {{B}_{net}}=\sqrt{{{({{B}_{H}}-{{B}_{wire}})}^{2}}+{{B}_{V}}^{2}} \\
& {{B}_{net}}=\sqrt{{{(0.32-0.2)}^{2}}+{{0.22}^{2}}} \\
& \therefore {{B}_{net}}=0.25G \\
\end{align}$
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.
$\text{1T=1}{{\text{0}}^{\text{-6}}}\text{G}$. The direction of the magnetic field produced by a current flowing through a straight wire is given by right hand thumb rule.
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