The current in the windings on a toroid is `2.0A`. There are 400 turns and the mean circumferential length is `40cm`. If the inside magnetic field is `1.0T,` the relative permeability is near to

Options

A. 200
B. 300
C. 400
D. 100

Answer

C

Explanation

For the toroid, `B=(mu_(0)mu_(R)Ni)/(2pir)`
`(because mu=mu_(0)muR` and for vacuum `mu_(R)=1 and mu=mu_(0))`
`therefore mu_(R)=(2pirB)/(mu_(0)Ni)` In this case `2pir=l=40xx10^-2m`
`=(40xx10^(-2)xx1)/(4pixx10^(-7)xx400xx2)`
`=(10^(4))/(8pi)=(10000)/(8xx3.14)=(10000)/(25)=400`