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CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 14 of 26
Marks: +1, -0
(a) An iron ring of relative permeability μr\mu_r has windings of insulated copper wire of nn turns per metre. When the current in the windings is II, find the expression for the magnetic field in the ring.
(b) The susceptibility of a magnetic material is 0.9853 . Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field.
Solution:  
(a) Expression for Ampere's circuital law
Derivation of magnetic field inside the ring
(b) Identification of the material
Drawing the modification of the field pattern
(a)From Ampere's circuital law, we have,
∮B⃗dl⃗=μ0μrIendosed\oint \vec{B} \vec{dl} = \mu_0 \mu_r I_{\text{endosed}} ......(i)
For the field inside the ring, we can write
∮B⃗dl⃗=∮B⃗dl⃗=B⋅2πr(r=radius of the ring)\oint \vec{B} \vec{dl} = \oint \vec{B} \vec{dl} = B \cdot 2\pi r \qquad (r = \text{radius of the ring})
Also, Ienclosed=(2πrn)I using equation (i)\text{Also, } I_{\text{enclosed}} = (2\pi r n) I \text{ using equation (i)}
∴B⋅2πr=μ0μr⋅(n⋅2πr)I\therefore B \cdot 2\pi r = \mu_0 \mu_r \cdot (n \cdot 2\pi r) I
∴B=μ0μrnI\therefore B = \mu_0 \mu_r n I
(b) The material is paramagnetic. The field pattern gets modified as shown in the figure below.
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