$A$ 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
    $8$
  • B
    $4$
  • C
    $2$
  • D
    $1$

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Similar Questions

$A$ steady current $I$ flows along an infinitely long hollow cylindrical conductor of radius $R$. This cylinder is placed coaxially inside an infinite solenoid of radius $2R$. The solenoid has $n$ turns per unit length and carries a steady current $I$. Consider a point $P$ at a distance $r$ from the common axis. The correct statement$(s)$ is (are) :
$(A)$ In the region $0 < r < R$,the magnetic field is non-zero.
$(B)$ In the region $R < r < 2R$,the magnetic field is along the common axis.
$(C)$ In the region $R < r < 2R$,the magnetic field is tangential to the circle of radius $r$,centered on the axis.
$(D)$ In the region $r > 2R$,the magnetic field is non-zero.

$A$ solenoid $2 \,m$ long and $4 \,cm$ in diameter has $4$ layers of windings of $1000$ turns each and carries a current of $5 \,A$. What is the magnetic field at its centre along the axis? $\left[\mu_0=4 \pi \times 10^{-7} \,Wb / Am\right]$

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $1000$ turns. It carries a current of $10 \ A$. What is the magnitude of the magnetic field inside the solenoid?

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved,the new value of the magnetic field is

The magnetic field intensity $(H)$ at the centre of a long solenoid having '$n$' turns per unit length and carrying a current $I$, when no material is kept in it is ($\mu_0$ = permeability of free space)

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