Question

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)?

Answer 1

The magnitude of the resultant of the magnetic field is
`4.11*10^(-5)\ T`

Step-by-step explanation:

Given that,

Current = 40 A

Magnetic field
`B=3.7*10^(-5)\ T`

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

`B'=(\mu_(0)I)/(2\pi r)`

Where, r = distance

I = current

Put the value into the formula

`B'=(4\pi*10^(-7)*20)/(2\pi*0.22)`

`B'=1.8*10^(-5)\ T`

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

Using formula of resultant

`B''=√(B^2+B'^2)`

Put the value into the formula

`B''=\sqrt{(3.7*10^(-5))^2+(1.8*10^(-5))^2}`

`B''=4.11*10^(-5)\ T`

Hence, The magnitude of the resultant of the magnetic field is
`4.11*10^(-5)\ T`