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julia-pushkina [17]
3 years ago
6

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)?
Physics
1 answer:
sladkih [1.3K]3 years ago
6 0

Answer:

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

Explanation:

Given that,

Current = 40 A

Magnetic field B=3.7\times10^{-5}\ T

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

B'=\dfrac{\mu_{0}I}{2\pi r}

Where, r = distance

I = current

Put the value into the formula

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

B'=1.8\times10^{-5}\ T

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

Using formula of resultant

B''=\sqrt{B^2+B'^2}

Put the value into the formula

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

B''=4.11\times10^{-5}\ T

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

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