Q.48. Two identical small bar magnets each of dipole moment J/T are placed at a center to center se

Q: What is the correct answer for JEE Main Physics Q48?

The correct answer is 12. Sol. $B_{net} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{5}$ $B_1 = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3}$ (axial field) $B_2 = \frac{\mu_0}{4\pi} \frac{2M}{(r/2)^3}$ (equatorial field) $B_{net} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{1^2 + 2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{5}$ Given $M = 3\sqrt{5}$ J/T, $r = 10$ cm $= 0.1$ m. $r/2 = 0.05$ m. $B_{net} =

Q: Q.48. Two identical small bar magnets each of dipole moment J/T are placed at a center to center separation of 10 cm, with their axes perpendicular to each other as shown in figure. The value of…

The correct answer is 12. Sol. $B_{net} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{5}$ $B_1 = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3}$ (axial field) $B_2 = \frac{\mu_0}{4\pi} \frac{2M}{(r/2)^3}$ (equatorial field) $B_{net} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{1^2 + 2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{5}$ Given $M = 3\sqrt{5}$ J/T, $r = 10$ cm $= 0.1$ m. $r/2 = 0.05$ m. $B_{net} =

Question

Q.48. Two identical small bar magnets each of dipole moment  J/T are placed at a center to center separation of 10 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  T. The value of  is [Figure: Two bar magnets, labeled S.N, are placed perpendicular to each other. The distance between their centers is 10 cm. Point P is midway between the centers of the magnets. B1 is the magnetic field due to the horizontal magnet at P, pointing downwards. B2 is the magnetic field due to the vertical magnet at P, pointing to the right.] ( Tm/A)

FirstInTest · Accepted Answer

Sol. $B_{net} = \sqrt{B_1^2 + B_2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{5}$ $B_1 = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3}$ (axial field) $B_2 = \frac{\mu_0}{4\pi} \frac{2M}{(r/2)^3}$ (equatorial field) $B_{net} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{1^2 + 2^2} = \frac{\mu_0}{4\pi} \frac{M}{(r/2)^3} \sqrt{5}$ Given $M = 3\sqrt{5}$ J/T, $r = 10$ cm $= 0.1$ m. $r/2 = 0.05$ m. $B_{net} = 10^{-7} 	imes \frac{3\sqrt{5}}{(0.05)^3} 	imes \sqrt{5} = 10^{-7} 	imes \frac{3 	imes 5}{125 	imes 10^{-6}} = 10^{-7} 	imes \frac{15}{125 	imes 10^{-6}} = 10^{-7} 	imes 120 	imes 10^{-3} = 12 	imes 10^{-7}$ T So $\alpha = 12$

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