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A single circular loop of wire with radius 0.02 cm carries a current of 8.0A.It is placed at the centre of a solenoid that has length 0.65 m,radius 0.080 mand 1300 turns. |
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Answer» The value of the current in the solenoid so that the magnetic field at the centre of the loop becomes zero,is EQUAL to `4.4 A`. Magnitude of `B_(solienoid)`=Magnitude of `B_("loop")` `mu_(0)NI=(mu_(0)I)/(2R)` here`n="Total no. of turn"/"Total length"=1300/0.65` `i=1/(2R)xx1/n=(8xx0.65)/(2xx0.02xx1300xx10^(-2))=10 A` For given condition: Total magnetic field at the centre of loop `=|B_("loop")|+|B_("solenoid")| :' |B_("loop")|=|B_("solenoid")|` `=2|B_("loop") =2xx(mu_(o)I)/(2R)` `=(2xx4pixx10^(-7)xx8)/(2xx0.02xx10^(-2))=16pixx10^(-3) T`. |
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