A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of magnitude 3.7 ✕ 10−5 T that is directed vertically downward, what is the resultant magnitude of the magnetic field 22 cm above the wire (in T)?

Respuesta :

Answer:

The magnitude of the resultant of the magnetic field is [tex]4.11\times10^{-5}\ T[/tex]

Explanation:

Given that,

Current = 40 A

Magnetic field [tex]B=3.7\times10^{-5}\ T[/tex]

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

[tex]B'=\dfrac{\mu_{0}I}{2\pi r}[/tex]

Where, r = distance

I = current

Put the value into the formula

[tex]B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}[/tex]

[tex]B'=1.8\times10^{-5}\ T[/tex]

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

[tex]B''=\sqrt{B^2+B'^2}[/tex]

Put the value into the formula

[tex]B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}[/tex]

[tex]B''=4.11\times10^{-5}\ T[/tex]

Hence, The magnitude of the resultant of the magnetic field is [tex]4.11\times10^{-5}\ T[/tex]