in this example we are given that a bar magnet
is zero point 1 meter long and its pole strength is 12 amp-ere meter. and here we are required
to find the magnetic induction at a point on its axis at a distance of zero point 2
meter from its centre. now in this situation if we. draw the picture. or we can see if
this is bar magnet. which is having length zero point 1 meter. and in this situation.
its. centre, is located at a distance of zero point 2 meter. from the point where we are
required to calculate the magnetic induction, we named this point say as p. now in this
situation if we consider this as r. and say, this is the north pole of magnet and this
is the south pole. then we know that due to north pole at point p the magnetic induction
will exist away from it and due to south pole it’ll exist toward it. so here we can write
that. net magnetic induction at point p. can be given as b-p is equal to, b-n minus b-s,
because north pole is closer to point p, so the magnetic induction due to it will be more.
and due to a magnetic pole. the magnetic induction is given as k m by the distance square so
due to the north pole. if we consider the pole strength to be m, as in a magnet the
2. poles have identical pole strength which we can denote by plus and minus or m-n and
m-s. here due to north pole we can directly write this is given as mu-not upon 4 pie.
m divided by the distance square if this distance is taken as l. then we can write it l minus
l by 2. so it is r minus l by 2 whole square. minus due to. the south pole it’ll be mu-not
upon 4 pie. m divided by the distance will b r plus l by 2 whole square. if we simplify
the result finally we are getting here mu-not m, over 4 pie we can take common. and if we
square these terms and we take l-c-m in numerator we’ll getting, this will be r-l plus r-l
it’ll be. 2 r-l divided by. r square minus l square by 4 whole square. and in this situation.
the final result if we simplify. then finally we’ll be getting here, if we just simplify
the result and substitute the numerical values you can see here. this 4 square 16 can be
taken in numerator so this will be. 32 mu-not m, r l. divided by 4 pie. 4 r square minus
l square. whole square. we substitute the values this magnetic field at point p will
be give by 32 mu-not upon 4 pie we can write as 10 to power minus 7. multiplied by. the
product of m and l, which is the dipole moment of this magnet which is already given at 12
amp-ere per meter we can take it 12. multiplied by, here. you can consider the value. of.
r to be zero point 2 meter. then divided by. if we substitute the values of r square and
l square over here. here we are getting it 4 multiplied by zero point 2 whole square
minus zero point 1 whole square. whole square. and if we simplify the expression directly
i can write down the final result which we’ll get is 3 point 4 into 10 to power minus 4
tesla. that will be the answer to this problem.
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