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Chapter 13: Problem 1977

Two iclentical short bar magnets, each having magnetic moment \(\mathrm{M}\) are placed a distance of \(2 \mathrm{~d}\) apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is. (a) $\sqrt{2}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (b) $\sqrt{3}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (c) $\sqrt{4}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (d) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

Short Answer

Expert verified
The short answer is: The magnetic induction at a point midway between the two identical short bar magnets is \(\sqrt{2}\left(\frac{\mu_{0}}{4\pi}\right)\left(\frac{M}{d^{3}}\right)\).

Step by step solution

01

Find the magnetic field due to one bar magnet

To find the magnetic field at the midpoint between the two bar magnets, we first need to find the magnetic field due to one bar magnet at that point. Let's call this bar magnet A. The magnetic moment of bar magnet A is given as M. Since the bar magnets are placed in a horizontal plane with a distance of 2d between them, the distance from the midpoint to bar magnet A is d. The formula for the magnetic field due to a magnetic dipole (bar magnet) at a distance r from it is given by: \[B = \frac{\mu_0}{4\pi}\frac{M}{r^3}\] Substituting the given values for M and r (d) in the formula, we get: \[B_A = \frac{\mu_0}{4\pi}\frac{M}{d^3}\]
02

Find the magnetic field due to the second bar magnet

Similarly, let's find the magnetic field due to the second bar magnet, B, at the midpoint. The magnetic moment of bar magnet B is also given as M and the distance from the midpoint to bar magnet B is also d. Using the formula for magnetic field due to magnetic dipole, we get: \[B_B = \frac{\mu_0}{4\pi}\frac{M}{d^3}\]
03

Combine the magnetic fields of the two bar magnets

Now, let's find the total magnetic induction at the midpoint by combining the magnetic fields due to both bar magnets A and B. Since the bar magnets are placed perpendicular to each other and the axes are perpendicular, their magnetic fields at the midpoint are at right angles to each other. We can use the Pythagorean theorem to combine these fields: \[B_{total} = \sqrt{B_A^2 + B_B^2}\] Substituting the values for B_A and B_B, we get: \[B_{total} = \sqrt{\left(\frac{\mu_0}{4\pi}\frac{M}{d^3}\right)^2 + \left(\frac{\mu_0}{4\pi}\frac{M}{d^3}\right)^2}\]
04

Simplify the expression

Now, we simplify the expression for B_total: \[B_{total} = \sqrt{2\left(\frac{\mu_0}{4\pi}\frac{M}{d^3}\right)^2}\] \[B_{total} = \sqrt{2}\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\] Comparing the expression of B_total with the given options, we find that the correct answer is: (a) \(\sqrt{2}\left(\frac{\mu_{0}}{4\pi}\right)\left(\frac{M}{d^{3}}\right)\)

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Most popular questions from this chapter

The time period of a freely suspended magnet is a 4 seconds. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be (a) \(4 \mathrm{sec}\) (b) \(2 \mathrm{sec}\) (c) \(0.5 \mathrm{sec}\) (d) \(0.25 \mathrm{sec}\)

The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)

A Galvanometer has a resistance \(\mathrm{G}\) and \(\mathrm{Q}\) current \(\mathrm{I}_{\mathrm{G}}\) flowing in it produces full scale deflection. \(\mathrm{S}_{1}\) is the value of the shunt which converts it into an ammeter of range 0 to \(\mathrm{I}\) and \(\mathrm{S}_{2}\) is the value of the shunt for the range 0 to \(2 \mathrm{I}\). The ratio $\left(\mathrm{S}_{1} / \mathrm{S}_{2}\right) \mathrm{is}$ (a) $\left[\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]$ (b) $(1 / 2)\left[\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]$ (c) 2 (d) 1

The magnetism of magnet is due to (a) The spin motion of electron (b) Earth (c) Pressure inside the earth core region (d) Cosmic rays

The magnetic induction at a point \(P\) which is at a distance \(4 \mathrm{~cm}\) from a long current carrying wire is \(10^{-8}\) tesla. The field of induction at a distance \(12 \mathrm{~cm}\) from the same current would be tesla. (a) \(3.33 \times 10^{-9}\) (b) \(1.11 \times 10^{-4}\) (c) \(3 \times 10^{-3}\) (d) \(9 \times 10^{-2}\)
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