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# A toroidal solenoid has 3000 turns and a mean radius of 10cm.It has a soft iron core of relative permeability 2000. Find the magnetic field in the core when a current of 1.0A is passed through the solenoid.(1)20T(2)12T(3)6T(4)3T

Last updated date: 13th Jul 2024
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Answer
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Hint: By knowing the values of permeability of the material of the soft iron core, number of turns, radius and current passing through the coil we can calculate the magnetic field. So to calculate the magnetic field first we have to calculate the permeability of the soft iron core which is the product of relative permeability, permeability of the material and permeability of free space. Thus substituting these values we will get the magnetic field of toroidal solenoid.
Magnetic field of toroidal solenoid is given by,
$B=\mu nI$
Where,
$n=\dfrac{N}{2\pi r}$
B is the magnetic field
N is the number of turns
r is the radius

Complete step-by-step solution:
Magnetic field of toroidal solenoid is given by,
$B=\mu nI$
Where,
$n=\dfrac{N}{2\pi r}$
B is the magnetic field
N is the number of turns
r is the radius
Give that N=3000
r=10cm
I=1.0A
${{\mu }_{r}}=2000$
First we have to calculate n.
$n=\dfrac{N}{2\pi r}$
$n=\dfrac{3000}{2\pi \times 10\times {{10}^{-2}}}$
Then we can calculate magnetic field using the equation,
$B=\mu nI$
Where,
$\mu ={{\mu }_{r}}{{\mu }_{0}}$
Thus B becomes,
$B={{\mu }_{r}}{{\mu }_{0}}nI$
Then by substituting the values we get,
$B=2000\times 4\pi \times {{10}^{-7}}\times \dfrac{3000}{2\pi \times 10\times {{10}^{-2}}}\times 1$
B=12T
Therefore option (2) is correct.

Note: The value of magnetic field B differs with the space. That is, the field inside and outside the toroidal is zero. For a toroid with closely bounded turns, the magnetic field has a constant magnitude. Using the right hand thumb rule we can determine the direction of the magnetic field.