
Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are ${I_1} = + 10A$,${I_2} = 13A$,${I_3} = + 10A$,${I_4} = + 7A$,${I_{_5}} = - 12A$ and ${I_6} = 18A$. The magnetic induction at a perpendicular distance of $10cm$ from the cable is (${\mu _0} = 4\pi \times {10^{ - 7}}Wb{A^{ - 1}}{m^{ - 1}}$)
A) $1.35\mu T$
B) $2.40\mu T$
C) $3.375\mu T$
D) $40.0\mu T$
Answer
233.1k+ views
Hint: In order to find the solution of the given question first of all we need to find the total current flowing in the wire. Then we need to directly apply the formula for the magnetic field induction and solve the required equation. After this we can conclude with the correct solution for the given question.
Complete step by step solution:
First of all let us find the total amount of current flowing in the cable. For that we need to find the sum of current flowing through all the six wires.
Therefore, we can write it as,
$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} + {I_6}$
$ \Rightarrow I = 10 - 13 + 10 + 7 - 12 + 18 = 20A$
The perpendicular distance from the cable is given as,$r = 10cm = \dfrac{{10}}{{100}}m = \dfrac{1}{{10}}m$
Let us write the formula for the magnetic field induction. It is given by,
$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}}$
Now, we know the formula for the magnetic field induction. So, we need to put the values in the equation to get the magnetic field induction.
Therefore, we can write the above equation after putting the values as,
$\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 20}}{{2\pi \times \dfrac{1}{{10}}}}$
$ \Rightarrow B = 40 \times {10^{ - 6}}T$
$\therefore B = 40\mu T$
Therefore, the required value of the magnetic induction of the cable at a perpendicular distance of $10cm$ is $40\mu T$.
Hence, option (D), is the correct choice for the given question.
Note: We should know this fact that magnetic induction due to a long straight wire at a perpendicular distance is given by, $B = \dfrac{{{\mu _0}I}}{{2\pi r}}$. We define magnetic induction when the production of an electromotive force is there across a conductor when the magnetic field changes. It can also be stated as the induced electromotive force in a loop when the magnetic flux is changed.
Complete step by step solution:
First of all let us find the total amount of current flowing in the cable. For that we need to find the sum of current flowing through all the six wires.
Therefore, we can write it as,
$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} + {I_6}$
$ \Rightarrow I = 10 - 13 + 10 + 7 - 12 + 18 = 20A$
The perpendicular distance from the cable is given as,$r = 10cm = \dfrac{{10}}{{100}}m = \dfrac{1}{{10}}m$
Let us write the formula for the magnetic field induction. It is given by,
$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}}$
Now, we know the formula for the magnetic field induction. So, we need to put the values in the equation to get the magnetic field induction.
Therefore, we can write the above equation after putting the values as,
$\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 20}}{{2\pi \times \dfrac{1}{{10}}}}$
$ \Rightarrow B = 40 \times {10^{ - 6}}T$
$\therefore B = 40\mu T$
Therefore, the required value of the magnetic induction of the cable at a perpendicular distance of $10cm$ is $40\mu T$.
Hence, option (D), is the correct choice for the given question.
Note: We should know this fact that magnetic induction due to a long straight wire at a perpendicular distance is given by, $B = \dfrac{{{\mu _0}I}}{{2\pi r}}$. We define magnetic induction when the production of an electromotive force is there across a conductor when the magnetic field changes. It can also be stated as the induced electromotive force in a loop when the magnetic flux is changed.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

