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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 \mathrm{~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 :
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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 $p \mu_0 \times 10^3$ 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 magnetic moment is zero
Reason: Magnetic field outside the volume of current carrying toroid is zero
A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as 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,
$\begin{aligned} & \oint B \cdot d l=\mu_0 \mathrm{NI} \\ & \text { or, } \\ & 2 \pi r \mathrm{~B}=\mu_0 \mathrm{NI} \\ & \text { or, } \\ & \mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \pi r}\end{aligned}$
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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