Please help! An air-core solenoid lias a length of 75.2 cm. a cross-sectional ar

Question

Please help!

An air-core solenoid lias a length of 75.2 cm. a cross-sectional area of 0.0039 in2, and contains 472 turns. The* current through the solenoid increases by 26.45 A. The permeability of free space is 1.25664 x 10-6 N/A2. 13y how much does the magnetic flux through the solenoid change? A rectangular wire loop with n turns Ls at rest in a uniform magnetic field B of magnitude 2 T that Ls directed into the page. The loop measures 6 cm by 8 cm, and the plane of the loop is perpendicular to the field, as shown. Niobium metal becomes h superconductor when cooled below 9 K. The permeability of free space is 1.25664 x 10"6 N/A2. If superconductivity is destroyed when the surface magnetic field exceeds 0.0628 T, determine the maximum current h 3.034 mm diameter niobium wire can carry and remain superconducting. Answer in units of A

Explanation / Answer

001) change in flux = muoNI2A/L - muoNI1A/L =  muoNA(I2-I1)/L

==> change in flux = 1.25664*10^-6*472*0.0939*26.45/0.752 = 0.00195896332 = 1958.96 muWb

002)intially Bo = muoI/2pia

after B = muoI/2pia +muoI/2pi*3a = muoI/2pia + 1/3 muoI/2pia = Bo+Bo/3 = 4/3 Bo

so correct option is 4.

003)flux = BA = 2*8*6/10000 = 0.0096 T.m2

so correct option is 4.

004) B = muoI/2pir

r = d/2 so B = muoI/pid

==> 0.0628 = 1.25664*10^-6*I/pi*3.034*10^-3

==> I = 0.0628*pi*3.034*10^-3/(1.25664*10^-6)

==> I = 476.33 A

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