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Question

A bar magnet of length 8 cm and having a pole strength of 1.0 Am is placed vertically on a horizontal table with its south pole on the table. A neutral point is found on the table at a distance of 6.0 cm north of the magnet. The horizontal component of earth's magnetic field is

A
44μT
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B
22μT
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C
11μT
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D
28μT
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Solution

The correct option is A 22μT
Given:
A bar magnet placed on a horizontal table having it's south pole on the table.


Solution:

Bn=μ4πm(d2+4l2)2 -------(i)

Bs=μ4πmd2 --------(II)

Now as at P there s null point so net BH will be equal .

BH=

μ4πmd2- μ4πmd(d2+4l2)32

now l=8/2=4cm, d=6cm,m=1Am

After calculate the final value wii 22μT.
So,correct opt: B


1998011_1315798_ans_262ab86d34524e16b4871b434002afb7.jpg

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