The magnetic field \vec{B} in acertain region is 0.128 ,and its direction is that of the z-axis in the figure.

sodni3

sodni3

Answered question

2021-01-13

The magnetic field B in acertain region is 0.128 ,and its direction is that of the z-axis in the figure.
image
Part A
What is the magnetic flux across the surface abcd in the figure?
Part B
What is the magnetic flux across the surface befc ?
Part C What is the magnetic flux across the surface aefd?
Part D
What is the net flux through all five surfaces that enclose the shaded volume?

Answer & Explanation

Faiza Fuller

Faiza Fuller

Skilled2021-01-14Added 108 answers

a) ϕB(abcd)=BA=0
b) ϕB(befc)=BA=(0.128T)(0.300 m)(0.300 m)=0.0115Wb
c) ϕB(aefd)=BA=BAcosϕ=35(0.128T)(0.500 m)(0.300 m)=+0.0115Wb
d) The net flux through the rest of the surface is zero since they are parallel to the x-axis so the total flux is the sum of all parts above, which is zero.
Jeffrey Jordon

Jeffrey Jordon

Expert2021-09-29Added 2605 answers

Step 1

When there is a change in magnetic flux then there is induced emf in the coil and the direction of the induced emf is such that it opposes the change and its direction is given by Lenz law. The magnetic flux (ϕ) can be given by,

ϕ=BA

where,

B is the magnetic field

A is the area

Step 2

a) The magnetic flux across the surface 'abcd' is,

ϕ=BAcosθ

ϕ=0.128T×(0.4m×0.3m)×cos90

since area vector is in negative x direction while magnetic field is in positive z direction.

ϕ=0

b) The magnitude flux across the surface 'befc' is,

ϕ=BAcosθ

ϕ=0.128T×(0.3m×0.3m)×cos180

since area vector is in negative z direction while magnetic field is in positive z direction.

ϕ=0.0115Wb

Step 3

c)

The magnetic flux across the surface 'aefd' is,

The area of surface 'aefd' is,

A=0.5×0.3=0.15m2

The angle can be calculated as,

cosθ=30cm50cm

ϕ=BAcosθ

ϕ=0.128T×(0.15)×35

ϕ=0.0115Wb

Step 4

d) The net flux through all five surfaces that enclose the shaded volume is given as,

ϕ=ϕabcd+ϕbefc+ϕaefd+ϕabe+ϕcdf

ϕ=00.0115+0.0115+0+0

ϕ=0

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