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The plane of a circular coil is at right angles to the magnetic meridian. If the number of turns in the coil equals 2, radius of the coil equals 0.078 m and current 2.8 A, then calculate the net magnetic field at the centre for clockwise and anticlockwise currents. Given : B_(u)=4 times 10^(-5)T. |
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Answer» Solution :Magnetic field at the cenre `=(mu_(0)/(4pi))((2pini)/r)tesla` i.e., `""B_(C)=((10^(-7))(2 times 3.142 times 2 times 2.8))/(0.078)` `""B_(C)=4.51 times 10^(-5)T.` Case (i) : For ANTI clockwise current `""B_(R)=B+B_(H)` `""B_(R)=(4.51+4) times 10^(-5)T=8.51 times 10^(-5)T` Case (ii) : For clockwise current `""B_(R)=B-B_(H)` `""=(4.51-4)times 10^(-5)` `""=0.51 times 10^(-5)T` 
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