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Find the magnitude of magnetic induction at a point 0.06m from the centre and along the axis of a circular coil carrying a current of 2 A. Also calculate the magnitude of magnetic induction at the centre of the coil. Given: Number of turns in the coil = 20 Mean radius of the coil = 0.05 m. |
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Answer» Solution :Given `I=2A, r=5 times 10^(-2)m, n=20, d=6 times 10^(-2)` We know that the magnetic field at the centre of the COIL, `""B=(mu_(0)/(4pi))((2pini)/r)T`. i.e., `""B=(10^(-7) times 2 times 3.142 times 20 times 2)/(5 times 10^(-2))=50.272 times 10^(-5)T` Magnetic fields at point `B^(')=(10^(-7)2pini)/(r(1+(x/r)^(2))^(3//2))` i.e., `""B^(')=(B)/((1+(x/r)^(2))^(3//2))` `""B^(')=(50.272 times 10^(-5))/(1+(0.06/0.05)^(2))^(3//2)=(50.272 times 10^(-5))/3.811` `""B^(')=13.19 times 10^(-5)T.` |
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