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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 magnetic susceptibility is negative for (a) Paramagnetic materials (b) Diamagnetic materials (c) Ferromagnetic materials (d) Paramagnetic and ferromagnetic materials

In each of the following questions, Match column-I and column-II and select the correct match out of the four given choices.\begin{tabular}{l|l} Column - I & Column - II \end{tabular} (A) Biot-savart's law (P) Direction of magnetic field induction (B) Right hand thumb rule (Q) Magnitude of magnetic field induction (C) Fleming's left hand rule (R) Direction of induced current (D) Fleming's right hand rule (S) Direction of force due to a mag. field (a) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (b) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$ (c) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (d) $\mathrm{A} \rightarrow \mathrm{P}: \mathrm{B} \rightarrow \mathrm{Q}: \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$

For the mag. field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be (a) \(0^{\circ}\) (b) \(90^{\circ}\) (c) \(180^{\circ}\) (d) \(45^{\circ}\)

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\( at an angle of \)60^{\circ}\( to the \)\mathrm{X}$ -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of \(r=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $4 \pi \times 10^{-7} \mathrm{sec}$
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