If the cross-sectional diameter (twice the radius) of the wire is 0.215 mm, and

Question

If the cross-sectional diameter (twice the radius) of the wire is 0.215 mm, and the wire is made of a material which has a resistivity of 1.5×10?6 ? m, how much power is dissipated in the wire loop? Answer in units of W.

Hint: dB/dt-b The circular loop of wire shown in the figure is placed in a spatially uniform magnetic field such that the plane of the circular loop is per pendicular to the direction for the magnetic field as shown in the figure. The magnetic field B(t) varies with time, with the time de- pendence given by s placed in a spatially unifor m ed B (t) a + bt. where a = 0.35 T and b 0.041 T/s. The acceleration due to gravity is 9.8 m/s2. r =8.1 cm radius B(t)

Explanation / Answer

Here ,

B = a + b*t

B = 0.35 + 0.041 * t

dB/dt = d/dt(0.35 + 0.041 * t )

dB/dt = 0.041 T/s

emf = change in flux/time

emf = 0.041 * pi * 0.081^2

emf = 8.45 *10^-4 V

radius of wire , R = p * L/A

R = 1.5 *10^-6 * 2pi * 0.081/(pi * (0.215 *10^-3)^2)

R = 5.26 Ohm

power dissipated = emf^2/R

power dissipated = (8.45 *10^-4)^2/5.26

power dissipated = 1.36 *10^-7 W

the power dissipated is 1.36 *10^-7 W

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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