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Magnetic Properties of Matter question

2013 · Shift 0 · Q60
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Magnetic Properties of Matter question

2013 · Shift 0 · Q60

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
Two short bar magnets of length 1cm1cm1cm each have magnetic moments 1.20Am21.20A{m^2}1.20Am2 and 1.00Am21.00A{m^2}1.00Am2 respectively. They are placed on a horizontal table parallel to each other with their NNN poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0cm.20.0cm.20.0cm. The value of the resultant horizontal magnetic induction at the mid-point OOO of the line joining their centres is close to ( \left( \, \right.( Horizontal component of earth's magnetic induction is 3.6×10.5Wb/m2)3.6 \times 10.5Wb/{m^2})3.6×10.5Wb/m2)
  1. A
    3.6×10.5  Wb/m23.6 \times 10.5\,\,Wb/{m^2}3.6×10.5Wb/m2
  2. B
    2.56×10.4  Wb/m22.56 \times 10.4\,\,Wb/{m^2}2.56×10.4Wb/m2
  3. C
    3.50×10.4  Wb/m23.50 \times 10.4\,\,Wb/{m^2}3.50×10.4Wb/m2
  4. D
    5.80×10.4 Wb/m25.80 \times 10.4\,Wb/{m^2}5.80×10.4Wb/m2
View written solutionFree

Correct answer: $1.6\TIMES 10^{-5}\,WB/M^2$, THE STORED ANSWER B SEEMS INCONSISTENT WITH THE STATED GEOMETRY AND STANDARD EQUATORIAL FIELD FORMULA.

  1. Given data
  • Magnetic moments: M1=1.20 A m2,M2=1.00 A m2M_1 = 1.20\,A\,m^2, \qquad M_2 = 1.00\,A\,m^2M1​=1.20Am2,M2​=1.00Am2
  • Separation between magnet centres: d=20.0 cm=0.20 md = 20.0\,cm = 0.20\,md=20.0cm=0.20m
  • Mid-point OOO is at distance r=d2=0.10 mr = \frac{d}{2} = 0.10\,mr=2d​=0.10m from each magnet.
  • Horizontal component of earth’s field: BH=3.6×10−5 TB_H = 3.6 \times 10^{-5}\,TBH​=3.6×10−5T

The magnets are parallel to each other and have a common magnetic equator. So point OOO lies on the equatorial line of each magnet.

  1. Magnetic field due to a short bar magnet on equatorial line

For a short bar magnet, magnetic induction at a point on the equatorial line is

Beq=μ04πMr3B_{eq} = \frac{\mu_0}{4\pi}\frac{M}{r^3}Beq​=4πμ0​​r3M​

where

  • MMM = magnetic moment,
  • rrr = distance from the centre of the magnet.

Thus,

B1=μ04πM1r3,B2=μ04πM2r3B_1 = \frac{\mu_0}{4\pi}\frac{M_1}{r^3}, \qquad B_2 = \frac{\mu_0}{4\pi}\frac{M_2}{r^3}B1​=4πμ0​​r3M1​​,B2​=4πμ0​​r3M2​​

Since both magnets are oriented in the same way, the equatorial fields at midpoint OOO are opposite in direction. Hence resultant due to the two magnets is the difference:

Bm=μ04πM1−M2r3B_m = \frac{\mu_0}{4\pi}\frac{M_1-M_2}{r^3}Bm​=4πμ0​​r3M1​−M2​​

Substitute values:

μ04π=10−7\frac{\mu_0}{4\pi} = 10^{-7}4πμ0​​=10−7

M1−M2=1.20−1.00=0.20 A m2M_1 - M_2 = 1.20 - 1.00 = 0.20\,A\,m^2M1​−M2​=1.20−1.00=0.20Am2

r3=(0.10)3=10−3 m3r^3 = (0.10)^3 = 10^{-3}\,m^3r3=(0.10)3=10−3m3

So,

Bm=10−7×0.2010−3B_m = 10^{-7}\times \frac{0.20}{10^{-3}}Bm​=10−7×10−30.20​

Bm=10−7×200B_m = 10^{-7}\times 200Bm​=10−7×200

Bm=2.0×10−5 TB_m = 2.0\times 10^{-5}\,TBm​=2.0×10−5T

  1. Direction of magnetic field due to magnets

At the midpoint on the common equatorial line, the resultant field due to the magnets is horizontal along the north-south direction.

Since the magnets have their north poles pointing toward the South, their magnetic moment vectors point toward geographic South. The stronger magnet dominates, so the net field due to magnets at OOO is toward geographic South.

Earth’s horizontal field BHB_HBH​ is toward geographic North.

Therefore these two fields are opposite in direction.

  1. Resultant horizontal magnetic induction

Hence,

Bnet=BH−BmB_{net} = B_H - B_mBnet​=BH​−Bm​

Bnet=3.6×10−5−2.0×10−5B_{net} = 3.6\times 10^{-5} - 2.0\times 10^{-5}Bnet​=3.6×10−5−2.0×10−5

Bnet=1.6×10−5 TB_{net} = 1.6\times 10^{-5}\,TBnet​=1.6×10−5T

  1. Compare with options

The computed value is

1.6×10−5 Wb/m21.6\times 10^{-5}\,Wb/m^21.6×10−5Wb/m2

But the listed options appear to have missing minus signs in exponents. Interpreting them as:

  • A: 3.6×10−53.6\times 10^{-5}3.6×10−5
  • B: 2.56×10−42.56\times 10^{-4}2.56×10−4
  • C: 3.50×10−43.50\times 10^{-4}3.50×10−4
  • D: 5.80×10−45.80\times 10^{-4}5.80×10−4

none matches directly.

However, if option B is intended to mean 2.56×10−5 T2.56\times 10^{-5}\,T2.56×10−5T, that would correspond to adding perpendicular components, which does not fit the geometry stated here.

So based on standard short magnet field formula and the given orientation, the resultant should be:

1.6×10−5 T\boxed{1.6\times 10^{-5}\,T}1.6×10−5T​

which does not match the stored answer.

  1. Conclusion

The correct physical result is:

1.6×10−5 Wb/m2\boxed{1.6\times 10^{-5}\,Wb/m^2}1.6×10−5Wb/m2​

So I do not agree with the stored correct answer BBB.

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