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Question: A telephone cable at a place has four long straight horizontal wires carrying the current of \(1A\) ...

A telephone cable at a place has four long straight horizontal wires carrying the current of 1A1A in the same direction east to west. The earth’s magnetic field at the place is 0.39G0.39G and the angle of dip is 3535^\circ the magnetic declination is zero. What are the resultant magnetic fields at the point 4cm4cmfrom the given cable?
A. 0.25G0.25G
B. 0.50G0.50G
C. 1.25G1.25G
D. 2.50G2.50G

Explanation

Solution

In this question, we have to find the resultant magnetic field of the telephone cable. Firstly, writing the horizontal and vertical component for the cable having current in the east to west direction. The magnetic field due to wire would also be in the direction opposite to the horizontal can be assumed by right hand thumb rule. After finding all such values we can find the resultant.

Complete Step by Step Answer:
Let us find the horizontal and the vertical components for the magnetic field. The magnetic field is given B=0.39GB = 0.39G and the dip is θ=35\theta = 35^\circ . Hence, horizontal component be given by
{B_H} = B\cos \theta \\\ \Rightarrow{B_H} = 0.39 \times \cos \left( {35} \right) \\\ \Rightarrow{B_H} = 0.32G \\\
Vertical component be given by
{B_V} = Bsin\theta \\\ \Rightarrow{B_V} = 0.39 \times sin\left( {35} \right) \\\ \Rightarrow{B_V} = 0.22G \\\
The telephone cable has total current I=1AI = 1Aflowing through it in east to west direction. The magnetic field acting on the wire by given by
Bwire=μ04π2Ir Bwire=1072×14×102{B_{wire}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2I}}{r} \\\ \Rightarrow{B_{wire}}= {10^{ - 7}}\dfrac{{2 \times 1}}{{4 \times {{10}^{ - 2}}}}
μ04π=107\because \dfrac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}} and the length of the cable is also given to be 4cm4cm.
Now, on solving Bwire=0.2G{B_{wire}} = 0.2G
Hence, net magnetic field be given by
Bnet=(BHBwire)2+BV2{B_{net}} = \sqrt {{{\left( {{B_H} - {B_{wire}}} \right)}^2} + {B_V}^2}
We had written the above equation as the Bh{B_h} and Bwire{B_{wire}} are acting in opposite directions hence, we firstly need to subtract them off and then we had taken their resultant. Now, substituting the values to find the answer.
{B_{net}} = \sqrt {{{\left( {0.32 - 0.2} \right)}^2} + {{\left( {0.22} \right)}^2}} \\\ \Rightarrow{B_{net}} = \sqrt {{{\left( {0.12} \right)}^2} + {{\left( {0.22} \right)}^2}} \\\ \therefore {B_{net}} = 0.25G \\\
Hence, the correct option is A.

Note: Choose the quantities of the system in the same units. Gauss is the cgs unit of magnetic field. We had calculated the direction by right hand thumb rule. The Right-Hand Thumb rule is also known as Maxwell’s corkscrew rule. If we consider ourselves driving a corkscrew in the direction of the current, then the direction of the corkscrew is in the direction of the magnetic field.