Question

Problem #2. A plastic core of length L and radius "a" has a long copper wire w nes. After N turns of wire, the outer radius of the coil is -4a". Te on top of itself, making the radius gradually bigger.) Assume the crwire. (20 pts) together tightly and uniformly. A current 1- is then run through the wire keeps wrapping r wires are packed A) Show a calculation of the magnitude of the magnetic field on the as' Law) core. (Hint: draw a rectangular cross-section of the coil, then use Ampere's Law.) B) Show a calculation of the magnitude of the magnetic field at a distance of 2a from the axis of the central plastic core. (Meaning, outside the plastic, inside the bulk of the copper wire.)

Answer 1

(A)M The rectangle cross-section ofthe coil-is-shown below: Ampere's loop C. R 4a A сoге The expression- for-the ampere's circuital-law-is, фв-а- д1я Ifthe wire consists-Nnumber of wires. Then the expression-is-written as-follows. В-dl - M,NI 1 Apply amperes circuital-law-for the loopABCD. we фВ-dl+ -get +фв-di-д,.м т B-dl B-dl дс DA B-dl 0 Since the length of the-loop and-magnetic-field From the figure,-AB-and-DA perpendicularto each other. So, the-dot productwill-bezero. are B-dl 0 4NI 0+ ф В-dl BC фВ-dl+фв-d-м, NI B DA The side BC- of the path-is- outside the solenoid. So, the magnetic-field outside the solenoid is equal to-zero-because-the length ofthe solenoid-is-greater than-the its diameter the magnetic field-outside-thesolenoidwill-be zero. В-dl -D01 BC Now В-dl -D p NI D4 вфа-д.мI DA Bl 4,NI

B= 14,NI We know that, the number of turns perunit-lengthis,T number of loops N length of coil В- д,n1 -1 Hence, the magnitude ofthe magnetic-field is 4nl (B)T Consider the ampere loop-inside-the-bulk-of the copper wire as follows:1 Ampere's loop C 4a A A core From the figure, the thickness ofthe copper loop is,1 - (4а-а) - За 1 Consider the thickness of-3a-copper loop-contains N-tuns. Now calculate-the number of turns for a thickness-cupper loop-as-follows: N a За (a) 3 Apply ampere's-law to the loop ABCD, В. dl - 4oNI 3 фв-аi. фВ-dl B- dl BC CD DA B-dl0 4NI 3 - dl+ 0+ BC CD B-dl B-d NI ac pA B1 (4nl)/= 4NI 3 - HoNI -(4)! 3 B .-(4n) 3 l)

The above negative sign indicates path-BC and-DA -are opposite to each other. Consider another loop,outside the plastic as follows.1 Ampere's loop D A 4a 2a core From the figure, the total-current enclosed-loop-is, 2NI (2a) 3a The ampere's law-for the loop ABCD-is, В dl + фв-di-2 NI 3 B-dl B-d CD DA В- di+0 - 24 B-dl 0+ cD фв-d+фв-а-21 NI 3 DA The side BC- of the path is- outside the solenoid. So, the magnetic field outside-the solenoid is-equal to zero. 2HNI ¢ B-dl 3 2H NI B= 3 - 2д, NI 31 2 ..2) From both-the equations the magnitude of the magnetic field is equal,-but-the negative sign shows both are in opposite direction. A