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:

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

see earlier