A rod with circular cross-section area $2\, cm ^2$ and length $40 cm$ is wound uniformly with $400$ turns of an insulated wire If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside windings is $4 \pi \times 10^{-6}\, Wb$ The relative permeability of the rod is(Given : Permeability of vacuum $\mu_0=4 \pi \times 10^{-7}\, NA ^{-2}$ )
A rod with circular cross-section area $2\, cm ^2$ and length $40 cm$ is wound uniformly with $400$ turns of an insulated wire If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside windings is $4 \pi \times 10^{-6}\, Wb$ The relative permeability of the rod is(Given : Permeability of vacuum $\mu_0=4 \pi \times 10^{-7}\, NA ^{-2}$ )
Show Hint
- $125$
- $12.5$
- $\frac{5}{16}$
- $\frac{32}{5}$
The Correct Option is A
Approach Solution - 1
ϕ = μrμ0\(\frac nl\)I × A
ϕ = 4π × 10−6 x 4π × 10−7 \(\frac {400}{0.40}\) x 0.4 x 2 x 10 -4
μr=125
So, the correct answer is (A): 125
Approach Solution -2
The correct option is (A): 125
By applying the Magnetic flux concept,
\(Φ = B.ACos \theta\)
As the area vector and magnetic field are tangential, both will be in a horizontal direction,
\(Φ = B.ACos0\degree\)
\(Φ = B.A\), where A is the cross-section area and
\(B = μ. n. i\), where again \(μ = μ_r. μ_0\)
μ = permeability of the
μr = relative permeability,
μ0 = permeability of vacuum,
n = no. of times,
i = current,
A = Area
\(Φ = μ_r . μ_0 . n . i . A\)
\(Φ=4\pi \times 10^{-6} \times 4\pi \times 10^{-7} \times \frac{400}{0.40} \times 0.4 \times 2 \times 10^{-4}\)
\(\rightarrow μ_r = \frac{100}{0.8}\)
\(\rightarrow μ_r = 125\)
Therefore, the Relative Permeability of the rod = 125
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Electromagnetic induction Questions
The magnitude of magnetic induction at the mid-point O due to the current arrangement shown in the figure is:
A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of earth in that region is 0.5 G and the angle of dip is \( 30^\circ \). The emf induced across the blades is \( N \pi \times 10^{-5} \, \text{V} \). The value of \( N \) is \( \_\_\_\_\_ \).
- Primary side of a transformer is connected to 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10 : 1. Load resistance connected to secondary side is 46 \(\Omega \). The power consumed in it is :
- A square loop of side \( 10 \, \text{cm} \) and resistance \( 0.7 \, \Omega \) is placed vertically in the east-west plane. A uniform magnetic field of \( 0.20 \, \text{T} \) is set up across the plane in the northeast direction. The magnetic field is decreased to zero in \( 1 \, \text{s} \) at a steady rate. Then, the magnitude of the induced emf is \( \sqrt{x} \times 10^{-3} \, \text{V} \). The value of \( x \) is _____.
- A horizontal straight wire 5 m long, extending from east to west, is falling freely at a right angle to the horizontal component of Earth's magnetic field \( 0.60 \times 10^{-4} \, \text{Wb/m}^2 \). The instantaneous value of emf induced in the wire when its velocity is \( 10 \, \text{m/s} \) is _______ \( \times 10^{-3} \, \text{V} \).
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
- When we place the conductor in a changing magnetic field.
- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
- t = time
Applications of Electromagnetic Induction
- Electromagnetic induction in AC generator
- Electrical Transformers
- Magnetic Flow Meter