Question

Type A.

1. (a) A rectangular frame of wire of 25 cm x 15 cm is placed
in a uniform electric field of strength 2 x 10-NC-, such
that the plane of the coil is normal to field. Find the electric
flux linked with the rectangular frame.
(6) Calculate the electric flux linked with the frame, when
it is converted into (1) a square and (ii) a circular frame.
(c) In which case is the electric flux maximum ?
O m ? C-1 (h) () SON m2 -1.​

Answer 1

a) 750W

b) 800W

1018.9W

c) the circular frame

Step-by-step explanation:

a) To find the electric flux you use the following formula:

`\Phi_E=EAcos\theta=EA`

where you have assumed that the plane of the loop and B are perpendicular between them.

By replacing you obtain:

`\Phi_E=(2*10^4N/C)(0.25m)(0.15m)=750W`

b) In order to calculate the electric flux in a square frame you first calculate the perimeter of the rectangular frame of wire:

P = 0.25m+0.25m+0.15m+0.15m=0.8m

A square with this length will have sides of 0.2m

Hence. you have that the electric flux is:

`\Phi_E=(2*10^4N/C)(0.2m)^2=800W`

for a circular frame you have that the radius is:

`s=2\pi r\\\\r=(s)/(2\pi)=(0.8m)/(2\pi)=0.127m`

Then, the electric flux will be:

`\Phi_E=(2*10^4N/C)(\pi)(0.127m)^2=1018.9W`

c) The electric flux is maximum in the circular frame.