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A toroid has a mean radius R equal to 20//pi cm, and a total of 400 turns of wire carrying a current of 2.0A. An aluminium ring at temperature 280 K inside the toroid provides the cure. (a) If the magnetization I is 4.8 xx 10^(-2) A/m, find the susceptibility of aluminium at 280 K. (b) If the temperature of the aluminium ring is raised to 320 K, what will be the magnetization ? |
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Answer» Solution :( a) The number of turns per unit length of the toroid is `n=(400)/(2piR)` The magnetic intensity H in the core is `H=ni=(400xx2.0A)/(2pixx20/pixx10^(-2)m)=2000A/m` The susceptibility is `CHI=I//H=(4.8xx10^(-2)A//m)/(2000A//m)=2.4xx10^(-5)` (B) The susceptibility X of a paramagnetic SUBSTANCE VARIES with absolute temperature as `chi = c //T`. Thus, `chi_(2)//chi_(1)=T_(1)//T_(2)` The susceptibility of aluminium at temperature 320 K is, THEREFORE, `chi=280/320xx2.4xx10^(-5)=2.1xx10^(-5)` Thus, the magnetization at 320 K is `I=chiH=2.1xx10^(-5)xx2000A//m=4.2xx10^(-2)A//m` |
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