Question

Two infinite wires 20 cm apart each carry a current of 3 A into the paper. d I I d/2 d/2 At a distance d 2 below their midpoint, what is the magnitude of the magnetic field? The acceleration of gravity is 9.8 m/s 2 and the permeability of free space is 1.2566 × 10−6 N/A 2 . Answer in units of µT.

Answer 1

Answer:

As electric current is carried in a cable, around it, a magnetic field is created. The lines of the magnetic fields form concentric circles around the wire. The direction of the magnetic field hinges on the direction of the current. It can be calculated by pointing the thumb of your right hand in the direction of the moment, using the "right hand law." The position of your curled fingers is in the magnetic field lines. The magnetic field magnitude depends on the sum of current, and the distance from the wire carrying the charge.

Explanation:

Determine the direction of vector B magnitude B: $B: B=\mu_{0} * 1 /(2 \pi r): r=d / 2 * \sqrt{2}$

$\cos \alpha=1 / 2 \Rightarrow \alpha=450$

Resultant magnitude strength: $B=2 B^{*} \cos \alpha=B=u_{0}^{*} 1 /(2 \pi r)=4.24^{*} 10^{-6} T=4.24 u T$ its direction is pointing to the left.

Note: Refer the image attached below