Acoil with circular cross section and 18 turns is rotating at a rate of 370 rpm (revolutions per minute) between the poles of a magnet. if the magnetic field strength is 0.7 t and peak voltage is 0.7 v, what is the radius of the coil (in cm)?

r= 2.1 cm

Explanation:

Given that

N= 18 turns

N= 370 rpm

B= 0.7 T

V= 0.7 V

The flux given as

Φ = B N A cosωt

B=Magnetic filed

N==Number of turns

A=Area

ω=Speed

t= time

The voltage V give as

Φ = B N A cosωt

V= BNA ω sinωt

The maximum value of voltage

V(max)= BNA ω

We know that ω=2 πf

ω = 38.74 rad/s

A=π r²

V(max)= BNA ω

0.7 = 0.7 x 18 x π r² x 38.74

r²=4.56 x 10⁻⁴ m²

r=0.021 m

r= 2.1 cm

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

i think it would be 5