Consider a tightly wound $100$ turn coil of radius $10 \,cm$ carrying a current of $2 \,A$. The magnitude of the magnetic field at the centre of the coil is

  • A
    $3.14 \times 10^{-4} \,T$
  • B
    $6.28 \times 10^{-4} \,T$
  • C
    $12.56 \times 10^{-4} \,T$
  • D
    $0$

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Similar Questions

For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

Two mutually perpendicular conductors carrying currents $I_1$ and $I_2$ lie in one plane. The locus of the point at which the magnetic induction is zero,is a

$A$ long curved conductor carries a current $I$. $A$ small current element of length $dl$ on the wire induces a magnetic field at a point away from the current element. If the position vector between the current element and the point is $\vec{r}$, making an angle $\theta$ with the current element, then the induced magnetic field density $d\vec{B}$ at the point is $(\mu_0 = \text{permeability of free space})$:

The magnetic field due to a current-carrying circular loop of radius $6 \ cm$ at a point on the axis at a distance of $8 \ cm$ from the centre is $216 \ \mu T$. Then the magnetic field at the centre of the ring is $\dots \ \mu T$.

The magnetic induction at a point $P$ which is distant $4 \, cm$ from a long current-carrying wire is $10^{-8} \, T$. The magnetic field induction at a distance $12 \, cm$ from the same current would be:

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