Two north poles of pole strengths $4 A - m$ and $16 A - m$ are separated by a distance of $20 c m$ . The position of the neutral point from $4 A - m$ pole is :

20 c m

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

18 c m

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

10 c m

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

6.67 c m

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

Open in App

Solution

The correct option is D $6.67 c m$
$m_{1} = 4 A m$
$m_{2} = 16 A m$
$B = \frac{μ_{o} m}{4 π d^{2}}$
$B_{1} = \frac{μ_{o} \times 4}{4 π \times (d^{'})^{2}}$
$B_{2} = \frac{μ_{o} \times 16}{4 π {(20 - d^{'})}^{2}}$
$B_{1} = B_{2}$
$\frac{1}{d^{'}} = \frac{2}{20 - d^{'}}$
$d^{'} = 6.67 c m .$

Suggest Corrections

Similar questions

Q. The north poles of pole strengths,

m

9 m

are kept apart by certain distance '

d

'. Find distance of null point from larger pole.

Q. Two magnetic poles of strengths

40 A m

and

10 A m

are separated by a distance of

20 c m

in air. The force between them if the distance is reduced to

10 c m

Q. The north poles of pole strengths m, 9m are kept apart by a certain distance. Find the distance of full point from larger pole.

The force between two similar poles of strengths $10 A m$ and $40 A m$ separated by a distance of $20 c m$ is

Two magnetic poles of pole strengths $324 m A - m$ and $400 m A - m$ . are kept at a distance of $10 c m$ in air. The null point will be at a distance of $- ------- -$ ( in $c m$ ), on the line joining the two poles, from the weak pole if they are like poles :

Join BYJU'S Learning Program

Two north poles of pole strengths 4 A−m and 16 A−m are separated by a distance of 20 cm. The position of the neutral point from 4 A−m pole is :

The correct option is D 6.67 cmm1=4Amm2=16AmB=μom4πd2B1=μo×44π×(d′)2B2=μo×164π(20−d′)2B1=B21d′=220−d′d′=6.67cm.

Two north poles of pole strengths $4 A - m$ and $16 A - m$ are separated by a distance of $20 c m$ . The position of the neutral point from $4 A - m$ pole is :

The correct option is D $6.67 c m$
$m_{1} = 4 A m$
$m_{2} = 16 A m$
$B = \frac{μ_{o} m}{4 π d^{2}}$
$B_{1} = \frac{μ_{o} \times 4}{4 π \times (d^{'})^{2}}$
$B_{2} = \frac{μ_{o} \times 16}{4 π {(20 - d^{'})}^{2}}$
$B_{1} = B_{2}$
$\frac{1}{d^{'}} = \frac{2}{20 - d^{'}}$
$d^{'} = 6.67 c m .$