A transformer with toroidal core of permeability μ is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius r R, and neglecting any leakage flux, the best estimate for the mean radius R is

Option 4 : \(\frac{{\mu I{r^2}N_P^2\omega }}{{2V}}\)

Given a transformer with toroidal care of permeability μ.

radius of the core = r

Mean radius of the core = R

cross sectional area of the core (A) = π r^{2}

mean length of core (L) = 2πR

\({V_p} = {N_p}\frac{{d\phi }}{{dt}} = {N_p}\frac{d}{{dt}}\left( {\frac{{{N_p}{i_p}}}{\mathbb{R}}} \right)\)

\( = \frac{{N_p^2}}{\mathbb{R}}\frac{d}{{dt}}\left( {I\sin \omega t} \right)\)

\(= \frac{{N_p^2}}{\mathbb{R}} \times \omega I\cos \omega t\)

\(\Rightarrow V\cos \omega t = \frac{{N_p^2}}{\mathbb{R}}\omega I\cos \omega t\)

we know that, \(\mathbb{R} = \frac{l}{{\mu A}}\)

\( \Rightarrow V = \frac{{N_p^2\omega I}}{{\left( {\frac{L}{{\mu A}}} \right)}}\)

\( \Rightarrow V = \frac{{N_p^2\;\mu A\;\omega I}}{L} = \frac{{N_p^2\omega I\mu \pi {r^2}}}{{2\pi R}}\)

\(\Rightarrow R = \frac{{N_p^2WI\mu {r^2}}}{{2V}}\)

Option 4 : 200

__Concept:__

According to Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of flux multiplied by the number of turns.

\(E = -N\frac{dϕ}{dt}\)

The negative sign is due to Lenz's law.

__Calculation:__

Given voltage is 50 V, dϕ = 70 - 20 = 50 mWb, dt = 0.2 s

Total number of turns is:

\(N = \frac{|E|}{|d\phi|}|dt|\)

\(N=\frac{50}{50 \times 10^{-3}}0.2\)

**N = 200**

Option 3 : Ampere’s circuital law states that the line integral of H̅ about any closed path is exactly equal to the direct current enclosed by the path

Let us analyze every option:

**Analysis for option (a)**

- When no voltage is applied between P N still we can observe a potential at their junction.
- Therefore we can say that the electric field present in depletion regions is directed from [+ve] to [-ve] i.e. from n-region to the p-region.
- Holes move from p-region to n-region and electron moves from n-region to the p-region
- Therefore option (a) is wrong

**Analysis of option (b)**

\(D = {\varepsilon _0}E + P\;;\;\left\{ {\begin{array}{*{20}{c}} {D = electric\;field\;Density\;\left( {\frac{{C}}{{{m^2}}}} \right)}\\ {E = electric\;field\;intensity\;\left( {\frac{V}{m}} \right)}\\ {P = polarization} \end{array}} \right.\)

**Analysis of option (c)**

Ampere’s circuit law:

It states that the line integral of the tangential component of "H” around a closed path is the same as the net current “I_{enc}” enclosed by the path.

i.e.\(\oint \vec H \cdot d\ell = {I_{enc}}\)

Note:

Observe that maxwell’s third equation obtained from Ampere’s law only:

We know, \({I_{enc}} = \oint \vec H \cdot d\ell = \mathop \smallint \nolimits_s \left( {\nabla \times \vec H} \right) \cdot ds\) ---(i)

But,

\({I_{enc}} = \mathop \smallint \nolimits_s \vec J \cdot ds\) ---(ii)

∴ \(\nabla \times \vec H = \vec J\)

option (c) is correct

Option 3 : 139 nJ/m^{3}

**Concept:**

The energy per unit volume, which is the instantaneous energy density uB associated with a magnetic field is given by:

\(\Rightarrow {u_B} = \frac{1}{2}\frac{{{B^2}}}{{{\mu _o}}} = \frac{1}{2}( \mu H^2)\)

**Analysis:**

εr = 1, μr = 1

Total magnetic energy density:

\(=\frac{1}{2} (\mu H^2)\)

Electric field intensity is given as:

E = 100 √ π V/m

The intrinsic impedance is given as:

\(\eta = \frac{E}{H}=377~ \Omega \)

\(H = \frac{100 \sqrt {\pi}}{377} = 0.470 ~A/m\)

Energy density is given as:

\(\frac{1}{2} \mu_0 H = \frac{1}{2}{4\pi \times 10^{-7}} \times (0.470)^2\)

= 139 nJ/m^{3}

Option 3 : \(\frac{I~r}{2πR^2}\)

**Concept:**

The magnetic field inside (for r < a) solid conducting cylindrical wire carrying current I is given by Amperes law

\(\begin{array}{l} \oint H.dl = {I_{enclosed}}\\ H = \frac{{Ir}}{{2\pi {a^2}}} \end{array}\)

For r = a

\(H = \frac{I}{{2\pi a}}\)

**Analysis:**

radius = R

distance from axis = r

∵ r < R

\(\begin{array}{l} \oint H.dl = {I_{enclosed}}\\ H = \frac{{Ir}}{{2\pi {R^2}}} \end{array}\)

Option 3 : ∇ × H = J

__Maxwell Equations:__

1) Modified Kirchhoff’s Current Law:

\(\nabla .\vec J + \frac{{\partial \rho }}{{\partial t}} = 0\)

J = Conduction Current density

2) Modified Ampere’s Law:

\(\nabla \times \vec H = \vec J + \frac{{\partial \vec D}}{{\partial t}}\)

Where \(\frac{{\partial \vec D}}{{\partial t}}\) = Displacement current density

3) Faraday’s Law:

\(\nabla .\vec E = - \frac{{\partial \vec B}}{{\partial t}}\)

4) Gauss Law:

\(\nabla .\vec D = \rho \)

Maxwell's Equations for time-varying fields is as shown:

Differential form |
Integral form |
Name |

\(\nabla \times E = - \frac{{\partial B}}{{\partial t}}\) |
\(\mathop \oint \nolimits_L^{} E.dl = - \frac{\partial }{{\partial t}}\mathop \smallint \nolimits_S^{} B.d S\) |
Faraday’s law of electromagnetic induction |

\(\nabla \times H =J+ \frac{{\partial D}}{{\partial t}}\) |
\(\mathop \oint \nolimits_L^{} H.dl = \mathop \smallint \nolimits_S^{} (J+\frac{{\partial D}}{{\partial t}}).dS\) |
Ampere’s circuital law |

∇ . D = ρv |
\(\mathop \oint \nolimits_S^{} D.dS = \mathop \smallint \nolimits_v^{} \rho_v.dV\) |
Gauss’ law |

∇ . B = 0 |
\(\mathop \oint \nolimits_S^{} B.dS = 0\) |
Gauss’ law of Magnetostatics (non-existence of magnetic monopole) |

Option 1 : 6.28 × 10^{-1}T

__CONCEPT:__

- Solenoid: A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.
- The uniform magnetic field is produced when an electric current is passed through it.

- The
**magnetic field inside a solenoid is proportional to the applied current and the number of turns per unit length**. - The magnetic field inside a solenoid does not depend on the diameter of the solenoid.
**The field inside is constant.**

B = μ_{0}N I

where N is no. of turns per unit length, I is current in the solenoid and μ_{0} is the permeability of free space.

**EXPLANATION:**

Given that I = 5 A

length of solenoid = 0.4 cm = 0.004 m

no. of turns per unit length N = 400 / 0.004 = 10^{5}

constant μ_{0} = 4π × 10^{-7}

** **Mgnetic field inside = μ_{0} N I = 4π × 10-7 × 10^{5 }× 5

The magnetic field inside the solenoid = **6.28 × 10-1 T**

- The magnetic field outside a solenoid is zero.

A transformer with toroidal core of permeability μ is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius r R, and neglecting any leakage flux, the best estimate for the mean radius R is

Option 4 : \(\frac{{\mu I{r^2}N_P^2\omega }}{{2V}}\)

Given a transformer with toroidal care of permeability μ.

radius of the core = r

Mean radius of the core = R

cross sectional area of the core (A) = π r^{2}

mean length of core (L) = 2πR

\({V_p} = {N_p}\frac{{d\phi }}{{dt}} = {N_p}\frac{d}{{dt}}\left( {\frac{{{N_p}{i_p}}}{\mathbb{R}}} \right)\)

\( = \frac{{N_p^2}}{\mathbb{R}}\frac{d}{{dt}}\left( {I\sin \omega t} \right)\)

\(= \frac{{N_p^2}}{\mathbb{R}} \times \omega I\cos \omega t\)

\(\Rightarrow V\cos \omega t = \frac{{N_p^2}}{\mathbb{R}}\omega I\cos \omega t\)

we know that, \(\mathbb{R} = \frac{l}{{\mu A}}\)

\( \Rightarrow V = \frac{{N_p^2\omega I}}{{\left( {\frac{L}{{\mu A}}} \right)}}\)

\( \Rightarrow V = \frac{{N_p^2\;\mu A\;\omega I}}{L} = \frac{{N_p^2\omega I\mu \pi {r^2}}}{{2\pi R}}\)

\(\Rightarrow R = \frac{{N_p^2WI\mu {r^2}}}{{2V}}\)

The magnetic field due to a current-carrying conductor takes the form of

Option 1 : Concentric circles

__Magnetic field due to a straight current-carrying conductor:__

The direction of magnetic field lines can be determined using the Right-Hand Thumb Rule.

The Magnetic field lines around a straight conductor carrying current are concentric circles whose centers lie on the wire.

The magnetic field path is shown below.

__Characteristics of Magnetic field due to a straight current-carrying conductor:__

- It encircles the conductor.
- It lies in a plane perpendicular to the conductor.
- The reversal in direction of current flow reverses the direction of the field.
- The strength of the field is directly proportional to the magnitude of the current.
- The strength of the field at any point is inversely proportional to the distance of the point from the wire.

Option 4 : 200

__Concept:__

According to Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of flux multiplied by the number of turns.

\(E = -N\frac{dϕ}{dt}\)

The negative sign is due to Lenz's law.

__Calculation:__

Given voltage is 50 V, dϕ = 70 - 20 = 50 mWb, dt = 0.2 s

Total number of turns is:

\(N = \frac{|E|}{|d\phi|}|dt|\)

\(N=\frac{50}{50 \times 10^{-3}}0.2\)

**N = 200**