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Two identical circular oils of radius `0*2m` each having 30 turns are mounted coaxially `0*2m` apart. What is the magnetic field at the centre of each coil when a current of `0*6A` is passed through both the coils (a) in the same direction (b) in the opposite directions? |
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Answer» Here, `a=0*2m`, `n=30`, `x=0*2m`, `I=0*6A` Magnetic field at the centre of each coil due to its own current is `B_1=(mu_0nI)/(2a)=((4pixx10^-7)xx30xx0*6)/(2xx0*2)` `=5*65xx10^-5T` Magnetic field at the centre of one coil due to the current in the other coil is `B_2=(mu_0)/(4pi)(2pinIa^2)/((a^2+x^2)^(3//2)` `=(10^-7xx2pixx30xx0*6xx(0*2)^2)/([(0*2)^2+(0*2)^2]^(3//2))` `=2*05xx10^-5T` (a) When the currents are in the same direction the resultant magnetic field at the centre of each coil is `B=B_1+B_2=5*65xx10^-5+2*05xx10^-5` `=7*70xx10^-5T` (b) When the currents are in the opposite directions, the resultant magnetic field at the centre of each coil is `B=B_1-B_2=5*65xx10^-5-2*05xx10^-5` `=3*60xx10^-5T` |
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