There are $50$ turns of a wire in every $1$ $cm$ length of a long solenoid. If $4$ $A$ current is flowing in the solenoid,the approximate value of the magnetic field along its axis at an internal point and at one end will be respectively:

  • A
    $12.6 \times 10^{-3} \, Wb/m^2, \; 6.3 \times 10^{-3} \, Wb/m^2$
  • B
    $12.6 \times 10^{-3} \, Wb/m^2, \; 25.1 \times 10^{-3} \, Wb/m^2$
  • C
    $25.1 \times 10^{-3} \, Wb/m^2, \; 12.6 \times 10^{-3} \, Wb/m^2$
  • D
    $25.1 \times 10^{-5} \, Wb/m^2, \; 12.6 \times 10^{-5} \, Wb/m^2$

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Similar Questions

There are $50$ turns per $cm$ length in a very long solenoid. It carries a current of $2.5 \ A$. The magnetic field at its centre on the axis is . . . . . . $T$.

The relative permeability in a core of a solenoid is $400$. The windings of a solenoid are insulated from the core and carry a current of $2 \ A$. If the number of turns is $1000$ per meter,then the magnetic intensity inside the core of the solenoid is . . . . . . $\frac{A}{m}$.

$A$ particle of mass $1 \times 10^{-27} \ kg$ and charge $1 \times 10^{-16} \ C$ enters a uniform magnetic field within a solenoid at a speed of $1000 \ m/s$. The velocity vector makes an angle of $60^{\circ}$ with the axis of the solenoid. The solenoid has $5000$ turns along its length $L$ and carries a current of $5 \ A$. The number of revolutions the particle makes along the helical path within the solenoid by the time it emerges from the solenoid's opposite end is:

In a toroid,the magnetic field on the axis is to be calculated given the radius $r = 0.5 \, cm$,current $I = 1.5 \, A$,number of turns $N = 250$,and relative permeability $\mu_r = 700$. Find the magnetic field $B$ (in Tesla). (in $.5$)

$A$ toroid has a core of inner radius $r_1$ and outer radius $r_2$,around which $N$ turns of wire are wound. If the current in the wire is $I$,then the magnetic field inside the toroid is $(\mu_0 = \text{permeability of free space})$

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