Question

1A

Two wires with currents 2 A and 1 A are enclosed in a circular loop.
Another wire with current 3 A is situated outside the loop as shown. 2A
The ØB. → around the loop is:
1) uo
2 3uo
3) 6
4)2uo​

Answer 1

we know according to the amperes circuital law

magnetic field around a closed loop is uo times the total current enclosed by the loop.

then ye answer

is 3 u0

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Answer 2

Answer:

The $\vec{OB}$ around the loop is $\mu_{0}$.

Explanation:

According to Ampere's circuital law

Ampere’s law states that “The magnetic field produced by an electric current exists proportionate to the size of that electric current with a constant of proportionality equivalent to the permeability of free space.”

Amperes law states that the line integral of $\vec{B}$ about any closed path or circuit exists equivalent to $\mu_{0}$ times the total current crossing the area specified by the closed path provided the electric field inside the loop stays constant.

Therefore,

by using Ampere's circuital law, we get

$\int \vec{B} \cdot \overrightarrow{d l}=\mu_{0} i_{n e t}$

$=\mu_{0}(3-2)=\mu_{0}$

Since currents 2 A and 1 A are in the opposite direction.

The $\vec{OB}$ around the loop is $\mu_{0}$.

Therefore, the correct answer is option 1). $\mu_{0}$

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