Question

14-51 The solenoid used in an MRI scanner is 2 m long and 90 cm in diameter. It is wound froma copper wire 1.4 mm in diameter, with adjacent turns separated by an insulating layer of varnish with negligible thickness. See example 26.10 of your textbook. 4. Find the current that will produce a 0.5 Find the voltage that needs to be applied to generate this current. Resistivity of the copper is 1.75 x 10 m. Hint: find the length of the wire first 5. T magnetic field. a. 203 A b. 557 A с. 897 A d. 1234 A e. 2315A a. 15 kV b. 25 kV c. 35 kV d. 45 kV e. 55 kV

Answer 1

The length of the solenoid is, L = 2 m. And the diameter of the copper wire is, d = 1.4 mm. So, the no. of turns in the copper wire is

The length of the solenoid is, L = 2m. And the diameter of the copper wire is, d = 1.4 mm, So, the no. of turns in the copper wire is N = L/d = 2-1.4 times 10^-3 = 1428 So, from the magnetic field formula in the solenoid, we write B = mu_0 nI = mu_0 N.L I = 4 pi times 10^-7 times 1428/2 times I I = 2 times 0.5/4 pi times 10^-7 times 1428 = 557 A So, option (b) is correct. Length of the wire is l = N times pi D Where D is the diameter of a single turn of the solenoid. So, putting the Given values, we get l = 1428 times pi times 0.9 = 4037.57 m The cross sectional area of the wire is A = pi d^2/4 = pi times (1.4 times 10^-3)^2/4 = 1.539 times 10^-6 m^2 The resistance of the wire is R = rho l/A = 1.75 times 10^-8 times

So, from the magnetic field formula in the solenoid, we write

$\Rightarrow I=\frac {2 \times 0.5} {4\pi \times 10^{-7}\times 1428}=557~A$

So, option (b) is correct.

5)

Length of the wire is

$l=N\times \pi D$

Where D is the diameter of a single turn of the solenoid.

So, putting the given values, we get

The cross sectional area of the wire is

The resistance of the wire is

Voltage difference

$V=IR =557\times 45.9=25.566~kV$

So, option ( b) is correct.