InterviewSolution
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InterviewSolution
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The current loop abcde formed by two circular segments of radii `r_1(=4cm)` and `r_2(=6cm)` with common centre at O, carries a current `I(=2A)` as shown in figure. What is the magnetic field at the common centre O? What will be the value of magnetic field at O when `theta=90^@`? |
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Answer» Magnetic field at the centre O due to current through the whole current loop is `B=B_(ab)+B_(bc)+B_(cd)+B_(dea)` directed normally downwards `=0+(mu_0)/(4pi)I/r_1theta+0+(mu_0)/(4pi)I/r_1(2pi-theta)` `=(mu_0I)/(4pi)[theta/r_2+(2pi-theta)/(r_1)]` directed normally downwards If `theta=90^@=(pi//2)rad`, then `B=(mu_0I)/(4pi)[(pi//2)/(r_2)+((2pi-pi//2))/(r_1)]=(mu_0I)/(8pi)[1/r_2+3/r_1]` `=((4pixx10^-7)xx2)/(8pi)[100/6+100/4]` `=41*7xx10^-7T` |
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