The ratio of magnetic field at the centre of a current carrying coil of radius to the magnetic field at distance from the centre of the coil on its axis is . The value of is _______. [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of the coil on its axis is $\sqrt{x} : 1$ . The value of $x$ is _______. [2023]

Ans.
(8)

$Magnetic field at centre (B_{1}) = \frac{μ_{0} I}{2 r}$

$Magnetic field on axis = \frac{μ_{0} I r^{2}}{2 {(r^{2} + d^{2})}^{3 / 2}}$

Value of $d = r$ (given)

$B_{2} = \frac{μ_{0} I}{4 \sqrt{2} r}$

$\frac{B_{1}}{B_{2}} = \frac{μ_{0} I}{2 r} \times \frac{4 \sqrt{2} r}{μ_{0} I} = \frac{2 \sqrt{2}}{1} = \frac{\sqrt{8}}{1}$

$\Rightarrow x = 8$

Want Us to Email you About Special Offers & Updates?