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Chapter 13: Problem 1956

Magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(\mathrm{M}\) is given by (a) \(\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)\) (b) \(\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{2}\right)\) (c) \(\left(\mu_{0} / 2 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)\) (d) \(\left(\mu_{0} / 2 \pi\right)\left(\mathrm{M} / \mathrm{d}^{2}\right)\)

Short Answer

Expert verified
The magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(M\) is given by (a) \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\).

Step by step solution

01

Recall the formula for Magnetic Intensity

The formula for magnetic intensity \(H\) at a point located on the axis of a short bar magnet is given by: \[H = \left(\frac{\mu_0}{4\pi}\right)\left(\frac{2M}{d^3}\right)\] Where \(\mu_0\) is the permeability of free space, \(M\) is the magnetic moment of the bar magnet, and \(d\) is the distance from the point to the center of the magnet.
02

Compare the given options with the correct formula

Now, let's compare the correct formula for magnetic intensity with the given options: 1. Option (a): \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\) 2. Option (b): \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^2}\right)\) 3. Option (c): \(\left(\frac{\mu_0}{2\pi}\right)\left(\frac{M}{d^3}\right)\) 4. Option (d): \(\left(\frac{\mu_0}{2\pi}\right)\left(\frac{M}{d^2}\right)\) The correct formula for magnetic intensity is \(H = \left(\frac{\mu_0}{4\pi}\right)\left(\frac{2M}{d^3}\right)\). Comparing this formula, we see that it matches Option (a).
03

Conclusion

Based on our comparison, we can conclude that the magnetic intensity for an axial point due to a short bar magnet of magnetic moment \(M\) is given by: (a) \(\left(\frac{\mu_0}{4\pi}\right)\left(\frac{M}{d^3}\right)\)

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Most popular questions from this chapter

A current of a \(1 \mathrm{Amp}\) is passed through a straight wire of length 2 meter. The magnetic field at a point in air at a distance of 3 meters from either end of wire and lying on the axis of wire will be (a) \(\left(\mu_{0} / 2 \pi\right)\) (b) \(\left(\mu_{0} / 4 \pi\right)\) (c) \(\left(\mu_{0} / 8 \pi\right)\) (d) zero

Read the assertion and reason carefully to mark the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion, (C) If assertion is true but reason is false. (D) If the assertion and reason both the false. (E) If assertion is false but reason is true. Assertion: Diamagnetic materials can exhibit magnetism. Reason: Diamagnetic materials have permanent magnetic dipole moment.

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5\) meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\). The \(\mathrm{B}\) is (a) \(\overline{(2 / 3) \text { tes } 1 a}\) (b) \((3 / 2)\) tesla (c) \((20 / 3)\) tesla (d) (3/20) tesla

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A coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that \(\mathrm{B}^{-}\) is in plane of the coil. If due to a current \(\mathrm{I}\) in the triangle a torque \(\tau\) acts on it, the side 1 of the triangle is (a) \((2 / \sqrt{3})(\tau / \mathrm{BI})^{1 / 2}\) (b) \((2 / 3)(\tau / B I)\) (c) \(2[\tau /\\{\sqrt{(} 3) \mathrm{BI}\\}]^{1 / 2}\) (d) \((1 / \sqrt{3})(\tau / \mathrm{BI})\)
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