Acircular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux through the loop is a maximum. at this position, the electric flux is 7.50 × 105 n⋅m2/c. determine the magnitude of the electric field. a) 8.88 × 105 n/c b) 1.07 × 106 n/c c) 2.44 × 106 n/c d) 4.24 × 106 n/c e) 6.00 × 106 n/c

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

where

E is the electric field

A is the cross-sectional area

is the angle between the direction of the electric field and the normal to A

The flux is maximum when , so we are in this situation and therefore , so we can write

Here we have:

is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

And so, we can find the magnitude of the electric field:

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//