(A) Given: Length of solenoid $l = 0.4 \,m$, number of turns $N = 400$, current $I = 5 \,A$.The number of turns per unit length $n = \frac{N}{l} = \fr

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(A) Given: Length of solenoid $l = 0.4 \,m$, number of turns $N = 400$, current $I = 5 \,A$.The number of turns per unit length $n = \frac{N}{l} = \frac{400}{0.4} = 1000 \,m^{-1}$.The magnetic field inside a long solenoid is given by $B = \mu_0 n I$.Substituting the values: $B = (4\pi \times 10^{-7}) \times 1000 \times 5$.$B = 4 \times 3.14159 \times 10^{-7} \times 5000$.$B = 20 \times 3.14159 \times 10^{-4} = 62.83 \times 10^{-4} = 6.28 \times 10^{-3} \,T$.

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

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$A$ solenoid has a length of $0.4 \,m$, a radius of $1 \,cm$, and $400$ turns of wire. If a current of $5 \,A$ is passed through this solenoid, what is the magnetic field inside the solenoid?

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A
$6.28 \times 10^{-3} \,T$
B
$6.28 \times 10^{-2} \,T$
C
$6.28 \times 10^{-4} \,T$
D
$6.28 \,T$

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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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$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

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$A$ solenoid has a length of $0.4 \,m$, a radius of $1 \,cm$, and $400$ turns of wire. If a current of $5 \,A$ is passed through this solenoid, what is the magnetic field inside the solenoid?

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$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

$A$ 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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