Question

he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of the wires is I = 4.7 A. In the drawing dH = 0.19 m and dV = 0.41 m. Find the magnitudes of the net magnetic fields at points A and B.

Answer 1

The question pertains to calculating the magnetic field created by current-carrying wires and the forces acting on them, using principles of electromagnetism such as the Biot-Savart law and the right-hand rule.

Step-by-step explanation:

The question asked involves calculating the magnetic field at specific points near current-carrying wires. This topic falls under the subject of Physics, particularly in the study of electromagnetism. The calculations require an understanding of the Biot-Savart law and the superposition principle for magnetic fields. To calculate the total magnetic field at a point due to multiple sources, one must consider the contribution of each current-carrying wire separately, using the right-hand rule to determine the direction of the individual fields and then vectorially adding them to find the net field.

• For a long straight wire, the magnitude of the magnetic field at a distance 'r' from the wire is given by µ_0I/2πr, where µ_0 is the permeability of free space and I is the current.
• The direction of the magnetic field created by a straight wire can be determined using the right-hand rule, with the thumb in the direction of the current and the curl of the fingers showing the direction of the field.

In solving these problems, the angle between the direction of the current and the magnetic field is crucial for calculating magnetic forces, as are the concepts of field lines, force, and the magnetic force equation F = Iℓ × B, where I is the current, ℓ is the length of the wire within the magnetic field, and B is the magnetic field.

Answer 2

The magnitudes of the net magnetic fields at points A and B is 2.66 x
`10^(-6)` T

Step-by-step explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ
`_(0)`I/2πr

where

B = magnetic field (T)

μ
`_(0)` = 1.26 x
`10^(-6)` (N/
`A^(2)`)

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ
`_(0)`I/2πr

= (1.26 x
`10^(-6)`)(4.7)/(2π)(0.19)

= 4.96 x
`10^(-6)` T

BV = μ
`_(0)`I/2πr

= (1.26 x
`10^(-6)`)(4.7)/(2π)(0.41)

= 2.3 x
`10^(-6)` T

hence,

the net magnetic field = BH - BV

= 4.96 x
`10^(-6)` - 2.3 x
`10^(-6)`

= 2.66 x
`10^(-6)` T