(C) The magnetic field inside an ideal solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length.Given: Length $L = 0.5 \

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(C) The magnetic field inside an ideal solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length.Given: Length $L = 0.5 \ m$, Radius $R = 0.1 \ m$, Current $I = 2.5 \ A$.Total turns $N = 2 \times 100 = 200$.Number of turns per unit length $n = N / L = 200 / 0.5 = 400 \ turns/m$.The magnetic field $B = \mu_0 n I = (4 \pi \times 10^{-7}) \times 400 \times 2.5$.$B = (4 \pi \times 10^{-7}) \times 1000 = 4 \pi \times 10^{-4} \ T$.Since the point is $5 \ cm$ from the axis, it lies inside the solenoid (as $5 \ cm Thus, the magnetic field is $4 \pi \times 10^{-4} \ T$.

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

Medium

$A$ solenoid of length $50 \ cm$ and radius $10 \ cm$ has two closely wound layers of windings, each with $100$ turns. If a current of $2.5 \ A$ is passing through the windings, the magnetic field (in $10^{-4} \ T$) at a point $5 \ cm$ from the axis is:

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A
$2 \pi$
B
$31.4$
C
$4 \pi$
D
Zero

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Class 12

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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 has an iron core with an internal magnetic field of $10 \pi \text{ mT}$,when the current in the winding of $1500 \text{ turns/m}$ is $10 \text{ A}$. Determine the field due to magnetization $(\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1})$.

MediumTS EAMCET 2019

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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.

EasyAIIMS 2015

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Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.

MediumAIIMS 2008

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$A$ long solenoid with $20$ turns per $cm$ is made. To produce a magnetic field of $20$ $mT$ inside the solenoid, the necessary current will be approximately......$A$.

Medium

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$A$ current of $1/4\pi \ A$ is flowing through a toroid. It has $1000$ turns per meter. The value of the magnetic field (in $Wb/m^2$) along its axis is:

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$A$ solenoid of length $50 \ cm$ and radius $10 \ cm$ has two closely wound layers of windings, each with $100$ turns. If a current of $2.5 \ A$ is passing through the windings, the magnetic field (in $10^{-4} \ T$) at a point $5 \ cm$ from the axis is:

Similar Questions

$A$ toroid has an iron core with an internal magnetic field of $10 \pi \text{ mT}$,when the current in the winding of $1500 \text{ turns/m}$ is $10 \text{ A}$. Determine the field due to magnetization $(\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1})$.

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.

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.Reason: The magnetic field inside the solenoid is uniform.

$A$ long solenoid with $20$ turns per $cm$ is made. To produce a magnetic field of $20$ $mT$ inside the solenoid, the necessary current will be approximately......$A$.

$A$ current of $1/4\pi \ A$ is flowing through a toroid. It has $1000$ turns per meter. The value of the magnetic field (in $Wb/m^2$) along its axis is:

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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.

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.