Question
Question: Six very long insulated copper wires are bound together to form a cable. The currents carried by the...
Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are I1=+10A,I2=13A,I3=+10A,I4=+7A,I5=−12A and I6=18A. The magnetic induction at a perpendicular distance of 10cm from the cable is (μ0=4π×10−7WbA−1m−1)
A) 1.35μT
B) 2.40μT
C) 3.375μT
D) 40.0μT
Solution
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=I1+I2+I3+I4+I5+I6
⇒I=10−13+10+7−12+18=20A
The perpendicular distance from the cable is given as,r=10cm=10010m=101m
Let us write the formula for the magnetic field induction. It is given by,
⇒B=2πrμ0I
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,
⇒B=2π×1014π×10−7×20
⇒B=40×10−6T
∴B=40μT
Therefore, the required value of the magnetic induction of the cable at a perpendicular distance of 10cm is 40μT.
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=2πrμ0I. 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.