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Find magnetic flux density at a point on the axis of a long solenoid having 5000 turns/m when it carrying a current of 2 A

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To find the magnetic flux density (B) at a point on the axis of a long solenoid, we can use the formula: \[ B = \mu_0 \cdot n \cdot I \] where: - \( B \) is the magnetic flux density, - \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)), - \( n \) is the number of turns per unit length of the solenoid (in turns/m), ...
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Knowledge Check

  • The torque required to hold a small circular coil of 10 turns, area 1mm^(2) and carrying a current of ((21)/(44))A in the middle of a long solenoid of 10 ^(3)"turns"//"m" carrying a current of 2.5A, with its axis perpendicular to the axis of the solenoid is

    A
    `1.5xx10^(-6)Nm`
    B
    `1.5xx10^(-8)Nm`
    C
    `1.5xx10^(6)Nm`
    D
    `1.5xx10^(8)Nm`
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