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Chapter 13: Problem 1968

A small bar magnet has a magnetic moment $1.2 \mathrm{~A} \cdot \mathrm{m}^{2}\(. The magnetic field at a distance \)0.1 \mathrm{~m}$ on its axis will be tesla. (a) \(1.2 \times 10^{-4}\) (b) \(2.4 \times 10^{-4}\) (c) \(2.4 \times 10^{4}\) (d) \(1.2 \times 10^{4}\)

Short Answer

Expert verified
The magnetic field at a distance of \(0.1\,\mathrm{m}\) along the axis of the given small bar magnet is \(2.4 \times 10^{-4} \mathrm{T}\). Answer: (b) \(2.4 \times 10^{-4}\).

Step by step solution

01

Recall the formula for magnetic field due to a magnetic dipole

The magnetic field B due to a magnetic dipole at a point on its axis is given by the formula: \[B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}\] Where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (equal to \(4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\)), \(M\) is the magnetic moment, and \(r\) is the distance from the dipole along its axis.
02

Substitute the given values into the formula

Given values are \(M = 1.2 \,\mathrm{A} \cdot \mathrm{m}^2\) and \(r = 0.1\,\mathrm{m}\). Substitute these values into the formula for the magnetic field: \[B =\frac{\mu_0}{4\pi} \cdot \frac{2(1.2\,\mathrm{A} \cdot \mathrm{m}^2)}{(0.1\,\mathrm{m})^3}\]
03

Calculate the magnetic field

Using \(\mu_0 = 4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\), let's plug in the value and calculate the magnetic field: \[B = \frac{(4\pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A})}{4\pi} \cdot \frac{2(1.2\,\mathrm{A} \cdot \mathrm{m}^2)}{(0.1\,\mathrm{m})^3}\] Simplify the expression: \[B = 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A} \cdot \frac{2.4\,\mathrm{A} \cdot \mathrm{m}^2}{0.001\,\mathrm{m}^3}\] \[B = 2.4 \times 10^{-4} \mathrm{T}\]
04

Match the calculated value with the given options

We find that the magnetic field at a distance of \(0.1\,\mathrm{m}\) along the axis of the given small bar magnet is \(2.4 \times 10^{-4} \mathrm{T}\), which corresponds to option (b). Answer: (b) \(2.4 \times 10^{-4}\)

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Most popular questions from this chapter

If a magnet of pole strength \(\mathrm{m}\) is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be (a) \((\mathrm{m} / 4)\) (b) \((\mathrm{m} / 2)\) (c) \((\mathrm{m} / 8)\) (d) \(4 \mathrm{~m}\)

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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

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) \(45^{\circ}\)

Two thin long parallel wires separated by a distance \(\mathrm{y}\) are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on other is (a) \(\left[\left(\mu_{0} I^{2}\right) / y^{2}\right]\) (b) \(\left[\left(\mu_{o} I^{2}\right) /(2 \pi \mathrm{y})\right]\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi)\right](1 / y)\) (d) $\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)$
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