Magnetic Properties of Matter

JEE Main level, numerical answer, medium

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Question 1 of 57

Two identical small bar magnets each of dipole moment 353\sqrt{5} J/T are placed at a center to center separation of 1010 cm, with their axes perpendicular to each other as shown in figure. The value of magnetic field at the point P midway between the magnets is α×10−3\alpha \times 10^{-3} T. The value of α\alpha is ______.
(μ0=4π×10−7\mu_0=4\pi \times 10^{-7} Tm/A)

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Answer: 12

Let the magnetic moment of each magnet be M=35M = 3\sqrt{5} J/T.

The distance of point P from the center of each magnet is r=5r = 5 cm =0.05= 0.05 m.

For the first magnet (horizontal), point P lies on its axial line. The magnetic field at P due to the first magnet is:

B1=μ04π2Mr3B_1 = \dfrac{\mu_0}{4\pi} \dfrac{2M}{r^3}

The direction of B⃗1\vec{B}_1 is along the axis of the first magnet.

For the second magnet (vertical), point P lies on its equatorial line. The magnetic field at P due to the second magnet is:

B2=μ04πMr3B_2 = \dfrac{\mu_0}{4\pi} \dfrac{M}{r^3}

The direction of B⃗2\vec{B}_2 is parallel to the axis of the second magnet, which is perpendicular to the axis of the first magnet.

Since B⃗1\vec{B}_1 and B⃗2\vec{B}_2 are perpendicular to each other, the net magnetic field at P is:

B=B12+B22B = \sqrt{B_1^2 + B_2^2}

Substituting the expressions for B1B_1 and B2B_2:

B=(μ04π2Mr3)2+(μ04πMr3)2B = \sqrt{\left(\dfrac{\mu_0}{4\pi} \dfrac{2M}{r^3}\right)^2 + \left(\dfrac{\mu_0}{4\pi} \dfrac{M}{r^3}\right)^2}

B=μ04πMr322+12=μ04π5Mr3B = \dfrac{\mu_0}{4\pi} \dfrac{M}{r^3} \sqrt{2^2 + 1^2} = \dfrac{\mu_0}{4\pi} \dfrac{\sqrt{5}M}{r^3}

Substituting the given values μ04π=10−7\dfrac{\mu_0}{4\pi} = 10^{-7} T m/A, M=35M = 3\sqrt{5} J/T, and r=0.05r = 0.05 m:

B=10−7×5×35(0.05)3B = 10^{-7} \times \dfrac{\sqrt{5} \times 3\sqrt{5}}{(0.05)^3}

B=10−7×15125×10−6B = 10^{-7} \times \dfrac{15}{125 \times 10^{-6}}

B=15×10−1125B = \dfrac{15 \times 10^{-1}}{125}

B=0.1×0.12B = 0.1 \times 0.12 T

B=0.012B = 0.012 T

B=12×10−3B = 12 \times 10^{-3} T

Comparing this with the given value α×10−3\alpha \times 10^{-3} T, we get α=12\alpha = 12.

Answer: 1212

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