question_answer

A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate

(i) H (ii) B (iii) Magnetising current \[{{I}_{m}}.\]

Where symbols have their usual meanings.

Answer:

It is given here that, \[{{\mu }_{r}}=400,\,\,I=2A,\,\,n=1000\,\,{{m}^{-1}}\] (i) Magnetic intensity H is independent of the material of core and is given by \[H=\frac{{{B}_{0}}}{{{\mu }_{0}}}=nl=1000\times 2.0=2\times {{10}^{3}}\,\,A{{m}^{-1}}\] (ii) The magnetic field, \[B=\mu H={{\mu }_{0}}{{\mu }_{r}}H\] \[=4\pi \times {{10}^{-7}}\times 400\times 2\times {{10}^{3}}T\] \[=1.0\,\,T\] (iii) The magnetising current \[{{I}_{m}}\] is the additional current needs to be passed through the winding of solenoid in the absence of core, which would make B value same as in the presence of core, Thus, \[B={{\mu }_{0}}\cdot n\,\,(I+{{I}_{m}})\] \[\therefore \]      \[(I+{{I}_{m}})=\frac{B}{{{\mu }_{0}}n}=\frac{1.0}{4\pi \times {{10}^{-7}}\times 1000}\approx 796\] \[{{I}_{m}}=796-I\] \[=796-2=794\,A\]