1.

Consider a toroid with rectangular cross section, of inner radius a, outer radius be and height h, carrying n number of turns. Then the self-inductance of the toroidal coil when current I passing through the toroid is

Answer»

`(mu_(0)n^(2)h)/(2pi)In((b)/(a))`
`(mu_(0)nh)/(2pi)In((b)/(a))`
`(mu_(0)n^(2)h)/(2pi)In((a)/(b))`
`(mu_(0)nh)/(2pi)In((a)/(b))`

Solution :Given a TOROID with a rectangular cross-section of INNER readius a and outer radius b. Height of the solenoid =h
Magnetic field inside a rectangular toroid is given by
`B=(mu_(0)nI)/(2pir)""......(i)`

usind the INFINITESIMAL cross-sectinoal area element.
`dx=hdr`
`:.` Flux passing through the cross-section of toroid.
`phi=intB.dx=int_(a)^(b)(mu_(0)nI)/(2pir).(hdr)`
`phi=(mu_(0)nIh)/(2pi)int_(a)^(b)(1)/(r)dr`
`impliesphi=(mu_(0)nhI)/(2pi)[logr]_(a)^(b)`
`impliesphi=(mu_(0)nhI)/(2pi)(logb-loga)`
`phi=(mu_(0)nhI)/(2pi)In((b)/(a))`
Now, self inductance of rectangular toroid,
`L=(nphi)/(I)`
PUTTING the value of `phi` we get
`L=(mu^(2)nh)/(2pi)In((b)/(a))`


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