1

**Tip 1: Field due to a bar magnet**

**What will be the field around a bar magnet of pole strength m and length 2l?**

At an

**axial point**the field due to a bar magnet is given by the resultant of field due to North Pole, $B_{1} $ and field due to South Pole, $B_{2} $$B_{axial} =4πμ_{o} (r_{2}−l_{2})_{2}Mr $

For a small bar magnet, l<

$B_{axial} =4πμ_{o} (r)_{3}M $

where $M$ is the magnetic moment($M=m(2l)$)

At an equatorial point, the field is given by,

$B_{equi} =−4πμ_{o} (r_{2}+l_{2})_{3/2}M $

for small bar magnet, l<

$B_{equi} =−4πμ_{o} (r)_{3}M $

**Now, How will you find the field die to a bar magnet at any arbitrary point?**

Let's simplify the steps to find the field due to a bar magnet.

- Determine the distance r at which the field is supposed to be calculated.
**$r$ is the vector between the point P and the center of the bar magnet making an angle $θ$ with the magnet.** - Resolve magnet N-S into N'-S' and N"-S" so that we can assume P to be in axial position for N'-S' and in equatorial position for N"-S".
- The magnetic moment $M$ can be resolved as $Mcosθ$ along the axial line, N'-S'. Similarly, magnetic moment $M$ can be resolved as $Msinθ$ along the equatorial line of N"-S"
- Field due to N'-S' can be written as, $B_{1} =4πμ_{o} (r)_{3}2Mcosθ $

Field due to N"-S" can be written as, $B_{2} =−4πμ_{o} (r)_{3}Msinθ $

- Net field can then be found as a resultant of $B_{1}$ and $B_{2}$

$B=B_{1}+B_{2} =4πμ_{o} r_{3}M 4cos_{2}θ+sin_{2}θ $

A bar magnet has pole strength $1Am$ and is placed as shown in Fig. $24.19$(a). Find the magnetic field at $P$.