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Question Number 99905 by Rio Michael last updated on 23/Jun/20

Commented by Rio Michael last updated on 23/Jun/20

P and Q are two identical straight cables each of electrical resistance 20.0Ξ©. placed 4.8 cm apart in air. (a) find the current through each cable (b) draw the magnetic field pattern in the region between P and Q (c) determine the resultant magnetic flux density at the point R due to P and Q only.

$$\mathrm{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{two}\:\mathrm{identical}\:\mathrm{straight}\:\mathrm{cables}\:\mathrm{each}\:\mathrm{of}\:\mathrm{electrical}\:\mathrm{resistance}\:\mathrm{20}.\mathrm{0}\Omega. \\ $$$$\mathrm{placed}\:\mathrm{4}.\mathrm{8}\:\mathrm{cm}\:\mathrm{apart}\:\mathrm{in}\:\mathrm{air}.\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{current}\:\mathrm{through}\:\mathrm{each}\:\mathrm{cable} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{draw}\:\mathrm{the}\:\mathrm{magnetic}\:\mathrm{field}\:\mathrm{pattern}\:\mathrm{in}\:\mathrm{the}\:\mathrm{region}\:\mathrm{between}\:{P}\:\mathrm{and}\:{Q} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{determine}\:\mathrm{the}\:\mathrm{resultant}\:\mathrm{magnetic}\:\mathrm{flux}\:\mathrm{density}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:{R}\:\mathrm{due}\:\mathrm{to} \\ $$$$\mathrm{P}\:\mathrm{and}\:{Q}\:\mathrm{only}. \\ $$

Commented by smridha last updated on 24/Jun/20

current through each cables I=((12)/(20))=0.6A(parallel combination) the direcrion of current flow through the each conductor(p and q) is same so the resultant mangnetic flux density is at R is B=[((𝛍_0 I)/(2𝛑(1.8Γ—10^(βˆ’2) )))+((𝛍_0 I)/(2𝛑(3Γ—10^(βˆ’2) )))]T =((4𝛑×10^(βˆ’7) Γ—0.6Γ—4.8Γ—10^(βˆ’2) )/(2𝛑×5.4Γ—10^(βˆ’4) )) =1.0666Γ—10^(βˆ’5) T

$$\boldsymbol{{current}}\:\boldsymbol{{through}}\:\boldsymbol{{each}}\:\boldsymbol{{cables}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{12}}{\mathrm{20}}=\mathrm{0}.\mathrm{6}{A}\left({parall}\boldsymbol{{el}}\:\boldsymbol{{combination}}\right) \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{direcrion}}\:\boldsymbol{{of}}\:\boldsymbol{{current}}\:\boldsymbol{{flow}} \\ $$$$\boldsymbol{{through}}\:\boldsymbol{{the}}\:\boldsymbol{{each}}\:\boldsymbol{{conductor}}\left(\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{q}}\right) \\ $$$$\boldsymbol{{is}}\:\boldsymbol{{same}}\:\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{resultant}}\:\boldsymbol{{mangnetic}} \\ $$$$\boldsymbol{{flux}}\:\boldsymbol{{density}}\:\boldsymbol{{is}}\:\boldsymbol{{at}}\:\boldsymbol{{R}}\:{is} \\ $$$$\:\:\boldsymbol{{B}}=\left[\frac{\boldsymbol{\mu}_{\mathrm{0}} \boldsymbol{{I}}}{\mathrm{2}\boldsymbol{\pi}\left(\mathrm{1}.\mathrm{8}Γ—\mathrm{10}^{βˆ’\mathrm{2}} \right)}+\frac{\boldsymbol{\mu}_{\mathrm{0}} \boldsymbol{{I}}}{\mathrm{2}\boldsymbol{\pi}\left(\mathrm{3}Γ—\mathrm{10}^{βˆ’\mathrm{2}} \right)}\right]\boldsymbol{{T}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}\boldsymbol{\pi}Γ—\mathrm{10}^{βˆ’\mathrm{7}} Γ—\mathrm{0}.\mathrm{6}Γ—\mathrm{4}.\mathrm{8}Γ—\mathrm{10}^{βˆ’\mathrm{2}} }{\mathrm{2}\boldsymbol{\pi}Γ—\mathrm{5}.\mathrm{4}Γ—\mathrm{10}^{βˆ’\mathrm{4}} } \\ $$$$=\mathrm{1}.\mathrm{0666}Γ—\mathrm{10}^{βˆ’\mathrm{5}} \boldsymbol{{T}} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 24/Jun/20

Brilliant sir..thank you. sir if iβ€²m to draw the magnetic pattern will it be like this? β†Ί β†Ί

$$\mathrm{Brilliant}\:\mathrm{sir}..\mathrm{thank}\:\mathrm{you}.\:\mathrm{sir}\:\mathrm{if} \\ $$$$\mathrm{i}'\mathrm{m}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{magnetic}\:\mathrm{pattern} \\ $$$$\mathrm{will}\:\mathrm{it}\:\mathrm{be}\:\mathrm{like}\:\mathrm{this}? \\ $$$$\:\:\circlearrowleft\:\:\:\:\:\:\:\circlearrowleft \\ $$

Commented by smridha last updated on 24/Jun/20

yeah ofcourse just simple as Ampearβ€²s circuital Law...take a closed loop around the wire and apply the right hand thumb rule..thatβ€²s easy for you..

$$\boldsymbol{{yeah}}\:\boldsymbol{{ofcourse}}\:\boldsymbol{{just}}\:\boldsymbol{{simple}} \\ $$$$\boldsymbol{{as}}\:\boldsymbol{{Ampear}}'\boldsymbol{{s}}\:\boldsymbol{{circuital}}\:\boldsymbol{{L}}{aw}...\boldsymbol{{take}} \\ $$$$\boldsymbol{{a}}\:\boldsymbol{{closed}}\:\boldsymbol{{loop}}\:\boldsymbol{{around}}\:\boldsymbol{{the}}\:\boldsymbol{{wire}} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{apply}}\:\boldsymbol{{the}}\:\boldsymbol{{right}}\:\boldsymbol{{hand}}\:\boldsymbol{{thumb}} \\ $$$$\boldsymbol{{rule}}..\boldsymbol{{that}}'{s}\:\boldsymbol{{eas}}{y}\:\boldsymbol{{for}}\:\boldsymbol{{you}}.. \\ $$

Commented by Rio Michael last updated on 24/Jun/20

thanks sir for the explanations

$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{the}\:\mathrm{explanations} \\ $$

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