Due to $10 \ A$ of current flowing in a circular coil of $10 \ cm$ radius,the magnetic field produced at its centre is $3.14 \times 10^{-3} \ Wb/m^2$. The number of turns in the coil will be:

  • A
    $5000$
  • B
    $100$
  • C
    $50$
  • D
    $25$

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Similar Questions

$A$ steady electric current is flowing through a cylindrical wire. Which of the following statements is/are correct?
$(a)$ The electric field at the axis of the wire is zero.
$(b)$ The magnetic field at the axis of the wire is zero.
$(c)$ The electric field in the vicinity of the wire is zero.
$(d)$ The magnetic field in the vicinity of the wire is zero.

In the above figure,the magnetic field at point $C$ will be:

Two concentric coils each of radius equal to $2\pi \, cm$ are placed at right angles to each other. $3 \, A$ and $4 \, A$ are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^2$ at the centre of the coils will be $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$.

The magnetic field induction at the centre of a circular coil of radius $5 \,cm$ carrying a current of $0.9 \,A$ is (in $SI$ units) (where $\varepsilon_0$ is the absolute permittivity of air in $SI$ units,and the velocity of light $c = 3 \times 10^8 \,ms^{-1}$):

$A$ wire has three different sections as shown in the figure. The magnitude of the magnetic field produced at the centre '$O$' of the semicircle by the three sections together is $(\mu_0 = \text{permeability of free space})$:

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