(B) The magnetic intensity $H$ inside a long solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the

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(B) The magnetic intensity $H$ inside a long solenoid is given by the formula $H = nI$,where $n$ is the number of turns per unit length and $I$ is the current flowing through the solenoid.Given:Total number of turns $N = 1200$Length of the solenoid $L = 2\,m$Current $I = 2\,A$First,calculate the number of turns per unit length $n$:$n = \frac{N}{L} = \frac{1200}{2} = 600\,m^{-1}$Now,calculate the magnetic intensity $H$:$H = nI = 600 \times 2 = 1200\,A m^{-1}$Therefore,$H = 1.2 \times 10^3\,A m^{-1}$.

Class 12 Physics Moving Charges and Magnetism Ampere’s circuital law and its application (Solenoid and Toroid)

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$A$ solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2\,m$ long and $0.2\,m$ in diameter. The magnetic intensity at the center of the solenoid when a current of $2\,A$ flows through it is:

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A
$2.4 \times 10^3\,A m^{-1}$
B
$1.2 \times 10^3\,A m^{-1}$
C
$1\,A m^{-1}$
D
$2.4 \times 10^{-3}\,A m^{-1}$

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Moving Charges and Magnetism

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Ampere’s circuital law and its application (Solenoid and Toroid)

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$A$ toroid is a long coil of wire wound over a circular core. If $r$ and $R$ are the radii of the coil and toroid respectively,the coefficient of self-induction of the toroid is (The magnetic field in it is uniform and $R >> r$). ($N =$ number of turns of the coil and $\mu_{0} =$ permeability of free space)

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$A$ solenoid of length $0.25 \ m$ has a radius of $1 \ cm$ and is made up of $500$ turns. It carries a current of $2.5 \ A$. What is the magnitude of the magnetic field inside the solenoid? (Take $\mu_0 = 4 \pi \times 10^{-7} \ SI$ units).

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Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.

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$A$ $90\, cm$ long solenoid has six layers of windings of $450\, turns$ each. If the diameter of the solenoid is $2.2\, cm$ and the current carried is $6\, A$, then the magnitude of the magnetic field inside the solenoid, near its centre is: (in $\pi\, G$)

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$A$ toroid has $500$ turns per metre length. If it carries a current of $2 \text{ A}$, the magnetic energy density inside the toroid is: (in $\text{ J/m}^3$)

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$A$ solenoid of $1200$ turns is wound uniformly in a single layer on a glass tube $2\,m$ long and $0.2\,m$ in diameter. The magnetic intensity at the center of the solenoid when a current of $2\,A$ flows through it is:

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$A$ solenoid of length $0.25 \ m$ has a radius of $1 \ cm$ and is made up of $500$ turns. It carries a current of $2.5 \ A$. What is the magnitude of the magnetic field inside the solenoid? (Take $\mu_0 = 4 \pi \times 10^{-7} \ SI$ units).

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.Reason : Magnetic field energy density is proportional to the square of the current.

$A$ $90\, cm$ long solenoid has six layers of windings of $450\, turns$ each. If the diameter of the solenoid is $2.2\, cm$ and the current carried is $6\, A$, then the magnitude of the magnetic field inside the solenoid, near its centre is: (in $\pi\, G$)

$A$ toroid has $500$ turns per metre length. If it carries a current of $2 \text{ A}$, the magnetic energy density inside the toroid is: (in $\text{ J/m}^3$)

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Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.