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. $44\mu T$
• B. $22\mu T$
• C. $11\mu T$
• D. $28\mu T$

Answer 1

Answer: (B)

$22\mu T$

Given:

A bar magnet placed on a horizontal table having it's south pole on the table.

Solution:

Bn=$\frac{\mu \cdot }{4\pi }\ast \frac{m}{(d^2+4l^2{)}^2}$ -------(i)

Bs=$\frac{\mu \cdot }{4\pi }\ast \frac{m}{d^2}$ --------(II)

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

BH=

$\frac{\mu \cdot }{4\pi }\ast \frac{m}{d^2}$- $\frac{\mu \cdot }{4\pi }\ast \frac{md}{(d^2+4l^2{)}^{\frac{3}{2}}}$

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

After calculate the final value wii 22$\mu$T.

So,correct opt: B