The radius of the current carrying coil is $R$ . The ratio of magnetic fields at the center of the coil to the axial point, which is $R \sqrt{3}$ distance away from the center of the coil is :

1 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 : 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $8 : 1$
REF. Image.
$B_{c} = \frac{M_{a} I}{2 R}$
$B_{a x i a l} = \frac{M_{0} I R^{2}}{2 (R^{2} + (\sqrt{3} R)^{2})^{3 / 2}} = \frac{M_{0} I R^{2}}{2 R^{3} (4)^{3 / 2}}$
$\frac{B_{C}}{B_{a x i a l}} = \frac{(M_{0} I .2 R)}{(M_{0} I / 2 R) (\frac{1}{8})}$
$\frac{B_{c}}{B_{a x i a l}} = \frac{8}{1}$

Suggest Corrections

Similar questions

Q. The radius of a current carrying coil is

^{'} R^{'}

. Then, the ratio of magnetic fields at the centre of the coil and at an axial point, which is

R \sqrt{3}

distance away from the centre of the coil is

Q. The magnetic field at the center of a circular coil of radius r carrying current is

B_{1}

. The field at the center of another coil of radius

\frac{r}{2}

carrying same current is

B_{2}

then ratio of

B_{1} / B_{2}

is :-

A circular coil of radius r carries current I. If another circular coil of radius 2r carries the same current, then the ratio of the magnetic fields at the centers of the circular is

Q. A circular current carrying coil has a radius

R

.The distance from the center of the coil on the axis where the magnetic induction will be (1/8)th of its value at the center of the coil, is :

Q. A circular coil of radius 3R is carrying current. The ratio of magnetic fields at centroid of coil and at a point distance 4R from center of coil (point is on the axis of coil)

Join BYJU'S Learning Program

The radius of the current carrying coil is R. The ratio of magnetic fields at the center of the coil to the axial point, which is R√3 distance away from the center of the coil is :

The correct option is B 8:1REF. Image.Bc=MaI2RBaxial=M0IR22(R2+(√3R)2)3/2=M0IR22R3(4)3/2BCBaxial=(M0I.2R)(M0I/2R)(18)BcBaxial=81

The radius of the current carrying coil is $R$ . The ratio of magnetic fields at the center of the coil to the axial point, which is $R \sqrt{3}$ distance away from the center of the coil is :