(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)

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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:

TS EAMCET 2025 All TS EAMCET

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

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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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$A$ 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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