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CBSE Class 12-science Answered

Show that the magnitude of magnetic induction at a point on the axis of a short bar magnet is twice the magnitude of magnetic induction at a point on the equator at the same distance.
Asked by kpbhake | 02 Mar, 2018, 09:49: AM
answered-by-expert Expert Answer
Let the length of barmagnet be 2l as shown in figure.  Let P be a point on the axis line at a distance d.
Magnetic induction at point P due to the poles of given barmagnet is given by

begin mathsize 12px style B subscript P space equals space fraction numerator mu subscript 0 over denominator 4 pi end fraction open parentheses m over open parentheses d minus l close parentheses squared space minus m over open parentheses d plus l close parentheses squared space close parentheses space equals fraction numerator mu subscript 0 over denominator 4 pi end fraction fraction numerator 4 l d space cross times m over denominator open parentheses d squared minus l squared close parentheses squared end fraction end style
where m is the pole strength of the given bar magnet. Let us assume d >> l and consider the magnetic moment = m×2l = M, then we can write the above relation as
begin mathsize 12px style B subscript P space equals space fraction numerator mu subscript 0 over denominator 4 pi end fraction fraction numerator 2 M over denominator d cubed end fraction end style.........................................(1)
Let Q be a point on the equatorial line at a distance d from bar-magnet as shown in figure. Magnetic induction at a point Q is given by
 
begin mathsize 12px style B subscript Q equals space 2 fraction numerator mu subscript 0 over denominator 4 pi end fraction fraction numerator m over denominator d squared plus l squared end fraction sin space theta space equals 2 fraction numerator mu subscript 0 over denominator 4 pi end fraction fraction numerator m over denominator d squared plus l squared end fraction fraction numerator l over denominator square root of d squared plus l squared end root end fraction space equals 2 fraction numerator mu subscript 0 over denominator 4 pi end fraction fraction numerator l space cross times m over denominator open parentheses d squared plus l squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent end fraction end style
Refer the diagram for the desciption of the induction vectors due to the poles and how the reultant vector is worked out.

Again by considering magnetic moment M = m×2l and d >> l, we can wite the above expression as

begin mathsize 12px style B subscript Q equals space fraction numerator mu subscript 0 over denominator 4 pi end fraction M over d cubed end style..........................................(2)
from (1) and (2), we can write BP = 2×BQ
 
Answered by Thiyagarajan K | 02 Mar, 2018, 01:16: PM
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