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Question: Two identical bar magnets, each having magnetic moment\[M\], are kept at a distance of \(2d\) with t...

Two identical bar magnets, each having magnetic momentMM, are kept at a distance of 2d2d with their axes perpendicular to each other in a horizontal plane. Find the magnetic induction midway between them.
A. (2)(\sqrt 2 ) μ04πMd3\dfrac{{{\mu _0}}}{{4\pi }}\dfrac{M}{{{d^3}}}
B. (3)(\sqrt 3 ) μ04πMd3\dfrac{{{\mu _0}}}{{4\pi }}\dfrac{M}{{{d^3}}}
C. μ04πMd3\dfrac{{{\mu _0}}}{{4\pi }}\dfrac{M}{{{d^3}}}
D. (5)(\sqrt 5 ) μ04πMd3\dfrac{{{\mu _0}}}{{4\pi }}\dfrac{M}{{{d^3}}}

Explanation

Solution

As, Bar magnets consist of North pole(N) and South pole(S) and the direction of magnetic induction is always from N to S. We will find magnetic induction due to both bar magnets at mid-point of their axes and will add them using vector algebra.

Formula used:
Since, B1B_1 due to first bar magnet produce magnetic induction in its axial line hence,
B1=μ04π×2Md3B_1 = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2M}}{{{d^3}}}
Where, MM is the magnetic moment of each given bar magnet and μ0{\mu _0} is the vacuum permeability.

Complete step by step answer:
Let us first draw the diagram, consider two bar magnets having both magnetic moments MM lying perpendicular to each other at a distance of 2d2d . We will measure this distance from mid-point of the axial bar magnet to the equatorial bar magnet.

Here, PP is the midpoint of two bars lying perpendicular to each other and At this point, B1B_1 is the magnetic induction due to the first bar magnet in the given direction and B2B_2 is the magnetic induction due to the second bar magnet. Since B1B_1 and B2B_2 are perpendicular to each other so net magnetic induction at point PP can be calculated as
Bp=(B1)2+(B2)2{B_p} = \sqrt {{{(B1)}^2} + {{(B2)}^2}} .........{Vector addition formula for two perpendicular vectors}
Since, B1B1 due to first bar magnet produce magnetic induction in its axial line hence,
B1=μ04π×2Md3B_1 = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2M}}{{{d^3}}}
B2B_2 Is the magnetic induction due to the second bar on its equatorial line which is given by?
B1=μ04π×Md3B_1 = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}.
Now, net magnetic induction Bp{B_p} is given by
Bp=μ04π×Md322+12{B_p} = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}\sqrt {{2^2} + {1^2}}
Bp=5\therefore {B_p} = \sqrt 5 μ04π×Md3\dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{d^3}}}.

Hence, correct option is D.

Note: Remember, Magnetic field lines always originate from North Pole (N) and enters to South Pole (S) which shows the direction of magnetic field from N to S but it’s important to note that inside the bar magnet the direction of magnetic induction is always from South Pole (S) to North Pole (N).