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Toroid - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 8 Questions around this concept.

Solve by difficulty

Average radius of toroid made on ring of non-magnetic material is 0.1 m it has 500 turns. If it carries 0.5 A current, then the magnetic field produced along its circular axis inside the toroid will be:

A toroidal solenoid has 3000 turns and a mean radius of 10 cm. It has a soft iron core of relative permeability 2000. What is the magnitude of the magnetic field in the core when a current of 1 A is passed through the solenoid?

A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury. When a current from the battery is started in the coil through the mercury :

A toroid of mean radius ' a ', cross-section radius ' r ' and a total number of turns N. it carries a current ' i '. The torque experienced by the toroid if a uniform magnetic field of strength  B  is applied.
 

A toroid having 1000 turns per meter and carrying a current 5 A the value of the magnetic field at point inside is pμ0×103 then the value of p is

For a toroid having n turns per unit length and carrying current i, the value of the magnetic field at the point inside the toroid is 

Direction:  In the following question, a statement of Assertion (A)  is followed by a statement of reason (R). Mark the correct choice as: 

Assertion: For a toroid, the magnetic moment is zero.

Reason: The magnetic field outside the volume of the current current-carrying toroid is zero.

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A toroid of n turns, mean radius R and cross-sectional radius a carry current I. It is placed on a horizontal table taken as an x-y plane. Its magnetic moment m.

Concepts Covered - 1

Toroid

Toroid -

If we try to bend a solenoid in the form of a ring then the obtained shape is a Toroid. So, a toroid can be considered a ring-shaped closed solenoid. Hence it is like an endless cylindrical solenoid. From the given figure we can understand Toroid much better.

                       

Now to obtain the magnetic field by a toroid, let us consider a toroid having N turns.

Here, we will now apply Ampere circuital law to calculate the magnetic field of a toroid. Suppose we have to find the magnetic field B at a point P inside the toroid as shown below in figure - 

                                       

Let us take an amperian loop which is a circle through point P and concentric inside the toroid. By symmetry, field will have equal magnitude at all points of this circle and this field is tangential to every point in the circle
Thus, 

                                                                          Bdl=μ0NI or, 2πr B=μ0NI or, B=μ0NI2πr

From the above result, B varies with r i.e. field B is not uniform over the cross-section of the core because as we increase 'r' the B varies.

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Toroid

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