Question

Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a uniform magnetic field that is directed at an angle of 31.0 ∘ above the horizontal. Part A What must the magnitude of the magnetic field be in order to produce a flux of 4.00×10−4 Wb through the surface? Express your answer with the appropriate units. BB = nothing nothing Request Answer Provide Feedback

Answer 1

Magnetic field, B = 0.88 T

Step-by-step explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is,
`A=8.82\ cm^2=0.000882\ m^2`

Angle between the uniform magnetic field and the horizontal,
`\theta=31`

Magnetic flux,
`\phi=4* 10^(-4)\ Wb`

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

`\phi=BA\ cos\theta`

`\theta` is the angle between magnetic field and the area

Here,
`\theta=90-31=59^(\circ)`

`B=(\phi)/(A\ cos\theta)`

`B=(4* 10^(-4))/(0.000882* cos(59))`

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.