Question

A 29.8 cm -diameter coil consists of 22 turns of circular copper wire 3.0 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.30×10⁻³ T/s .

A. Determine the current in the loop

B. Determine the rate at which thermal energy is produced.

Answer 1

Diameter of the coil, d = 29.8 cm = 0.298 m

Hence, radius of the coil, R = d/2 = 0.298/2 = 0.149 m

Therefore, EMF between ends of the coil, E = N*d(BA)/dt

where N = no. of turns, B = magnetic field and A = loop area

Put the values -

E = 22 * (8.30 x 10^(-3)) * [ π (0.149)^2 ] volt
= 12.73 x 10^(-3) volt

Now, radius of the copper wire, r = (3.0 x 10^-3) / 2 = 1.5 x 10^-3 m

So, resistance of coil = ρ * (l / A )

resistivity of copper = 1.77 x 10^(-8) ohm*m

length of the copper wire, l = NxCircumference of the wire = Nx2xpixR = (22x2x3.141x0.149) m

So, resistance of the coil = (1.77 x 10^(-8)) * (22x2x3.141x0.149)) / [3.141 x (1.5 x 10^(-3))^2]

= 5.16 x 10^(-2) Ω

(A) Current in the loop = EMF / resistance
= [ 12.73 x 10^(-3) ] / [5.16 x 10^(-2) ]
= 0.247 A

(B) Rate of thermal energy = emf x current / J cal/s
= [12.73 x 10^(-3)] * [0.247] / 4.2 cal/s
= 0.749 x 10^(-3) cal/s. (Answer)