Q. 174.2( 11 Votes )

# A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound.

If the current in the wire is 11 A, what is the magnetic field

(a) outside the toroid,

(b) inside the core of the toroid, and

(c) in the empty space surrounded by the toroid.

Answer :

The Toroid is same as a solenoid but is bound in a circular region as compared to straight cylindrical region in a solenoid the length of a toroid is given by L = 2πR, (Perimeter of a Circle)

where R is mean radius of cross section of toroid

= 25.5 cm = 0.255 m

here Total number of turns of toroid N = 3500,

no. of turns per unit length n = N/L = = 2475.74/m

the current flowing in Toroid I = 11 A

**The toroid has been shown in the following figure**

(a) Since Toroid is same as a solenoid so magnetic field outside the toroid = 0 T same as in case of solenoid.

__Explanation: if we apply Ampere Circuital Law__ __to find magnetic field B at small element dl due to a current carrying conductor with current I flowing through it and__ __is permittivity of free space, outside the solenoid or toroid, where there is no conductor/wire/current carrying loop, the current I = 0 so we get__ __So Integrating over length l we get Bl = 0 so magnetic field B = 0 i.e. at all places where there is no current carrying loop magnetic field B = 0__

(b) The magnetic field in the inner core of toroid is given by

B = 𝜇onI

Where n is no. of turns per unit length,

I is the current flowing in Toroid,

𝜇o is the permittivity of free space = TmA^{-1}

So putting all values, B = = T

i.e. Magnetic field inside core is 0.03 T.

(c) Again it’s the same case as of exterior of a solenoid as explained above so magnetic field is Zero, here also in the empty space between the toroid.

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