A 20 turn square coil has an area 19.0 cm2 and is rotated with an angular speed

Question

A 20 turn square coil has an area 19.0 cm2 and is rotated with an angular speed of 170.0 rad/s in a uniform 0.30 T magnetic field. The axis of the coil is perpendicular to the field at all times. If the coil forms a closed circuit with a series resistance of 50.0 , what is the instantaneous power being dissipated in the resistor when the plane of the coil makes an angle of 18.0° with the magnetic field?

V = N*B*A*w*sin(90 - 18.0) V = 20 * 0.30 * 19.0 * 10^-4 * 170.0 * sin(72) P=V^2/R=0.0679 V*A

But my answer is not correct.

E = BANsin = BANsin() = 0.30*19*10^-4 *20*170 * sin( 18) = 0.598875 V P =V^2 /R = 0.598875^2 / 50 = 0.00717302531 W

It is also incorrect

Explanation / Answer

V = N*B*A*w*sin(90-18) = 20*0.3*19*10^-4*170*sin(90-18) = 1.84 V

Instantaneous Power P = Vrms^2/R

Vrms = V/sqrt(2) = 1.84/1.414 = 1.3 V

Then Power P = Vrms^2/R = 1.3^2/50 = 0.0338 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.