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Two circular coils of mean radii 0.1 m and 0.5 m consisting of 5 turns and 10 turns respectively are arranged concentric to one another with their planes at right angles to each other. If a current of 2A is passed through each of them, calculate the magnitude of the resultant magnetic field at their common centre. |
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Answer» Solution : Given : `r_(1)=0.10 m, r_(2)=0.05m, n_(1)=5, n_(2)=10, I=2A` We know that, B at the centre of the circular conductor carrying current `""B=mu_(0)/(4PI)(2pinI)/r""tesla` Hence, `""B_(1)=(10^(-7) times 2 times 3.142 times 5 times 2)/(0.10)` `""B_(1)=6.284 times 10^(-5)T` Similarly, `""B_(2)=(10^(-7) times 2 times 3.142 times 10 times 2)/(0.05)` `""B_(2)=25.136 times 10^(-5)T` Resultant field, `""B=sqrt(B_(1)^(2)+B_(2)^(2))` `""=10^(-5)sqrt((284)^(2)+(25.136)^(2))` `""=10^(-5)sqrt(39.489 times 631.818)` `""=10^(-5)sqrt(671.307)` `""=10^(-5) times 25.91` Resultant field at the centre = `2.591 times 10^(-4)T` `""alpha=tan^(-1)(B_(2)/B_(1))` `""=tan^(-1)((25.136 times 10^(-5))/(6.284 times 10^(-5)))` `""a=tan^(-1)(4) " w.r.t "B_(1)` i.e., `""a=75^(@)58^(') " w.r.t "B_(1)` The direction of the resultant is `75^(@)58^(')` w.r.t. the direction of `B_(1)` 
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