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Home » 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°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
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°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
CBSE | Class 12 | Magnetism and Matter | NCERT | Physics
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°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Answer –

First it is important to decide the direction which would best represent the given situation.

We are given that

BH = B cos δ = 0.39 × cos 35o G

Therefore, BH = 0.32G

Here

BV = B sin δ = 0.39 × sin 35o G

Therefore, BV = 0.22G

The telephone cable is assumed to carry a total current of 4.0 A in the east-west direction. As a result, the magnetic field generated is 4.0 cm below.

    \[{{B}_{wire}}=\frac{{{\mu }_{0}}2I}{4\pi r}={{10}^{-7}}\times \frac{2\times 4}{4\times {{10}^{-2}}}\]

    \[{{B}_{wire}}=2\times {{10}^{-5}}T=0.2G\]

Now, the net magnetic field is given by –

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

    \[{{B}_{net}}=0.25G\]

0.25 G is the resultant magnetic field at points 4 cm below the cable.

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