Question

A 29 cm diameter coil consists of 22 turns of cylindrical copper wire 2.6 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 7.00

Answer 1

Note that

V = NA(dB/dt) [1]

where

N = number of turns (22)
A = cross sectional area
dB/dt = rate of change in the magnetic field (7.00E-3 T/s)

Here, we have circular cross section. Thus,

A = pi D^2 /4 [2]

where D = the diameter of the coil (0.29 m)

Thus, substituting [2] to [1],

V = pi N D^2 (dB/dt)/4 [3]

Now, note that

I = V/R

where I = current

R = resistance

Thus,

I = V/R = pi N D^2 (dB/dt)/4R [4]

Now, to get R, note that

R = rho(L)/a

where L = the total length of wire (22 turns * circumference = 20.04 m)

a = cross section area of the wire (pi *d^2 / 4 = 5.309E-6 m^2)

Thus,

I = V/R = pi N D^2 (dB/dt) / [4R] [4]

becomes

I = V/R = pi N D^2 a (dB/dt)/[ 4 (rho) L]

Plugging in the values,

I = pi N D^2 a (dB/dt)/[4 (rho) L]

=3.1416 (22) (0.29)^2 (5.309E - 6) (7.00E-3) / [4(1.72E-8)(20.04)]

= 0.157 A

= 157 mA [ANSWER]

DONE! It was nice working with you!