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Chapter 13: Problem 1896

Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment is (a) \(2: \pi\) (b) \(\pi: 2\) (c) \(\pi: 4\) (d) \(4: \pi\)

Short Answer

Expert verified
The ratio of the magnetic moment of the square to the circle is \(\pi: 2\), which corresponds to option (b).

Step by step solution

01

Calculate the magnetic moment of the square wire

Let's denote the length of the wire as L. Since the wire is shaped into a square, each side of the square will have a length of L/4. The area A_s of the square can be calculated as: \[A_s = (\frac{L}{4})^2\] Now, let's denote the current passing through the wire as I. The magnetic moment M_s of the square can be calculated as: \[M_s = I \cdot A_s\] By substituting the area of the square, we get: \[M_s = I \cdot (\frac{L}{4})^2\]
02

Calculate the magnetic moment of the circle wire

To find the magnetic moment of the circle wire, we first need to determine its radius. Since the circumference of the circle is equal to the length of the wire, we have: \[2\pi r = L\] Solving for the radius r, we get: \[r = \frac{L}{2\pi}\] Now, we can calculate the area A_c of the circle as: \[A_c = \pi r^2\] Substituting the radius, we get: \[A_c = \pi (\frac{L}{2\pi})^2\] The magnetic moment M_c of the circle can be calculated as: \[M_c = I \cdot A_c\] By substituting the area of the circle, we get: \[M_c = I \cdot \pi (\frac{L}{2\pi})^2\]
03

Calculate the ratio of the magnetic moments

We need to find the ratio of the magnetic moments of the square and the circle: \[\frac{M_s}{M_c} = \frac{I \cdot (\frac{L}{4})^2}{I \cdot \pi (\frac{L}{2\pi})^2}\] We can see that the current I and the length L will cancel out, so the ratio of the magnetic moments simplifies to: \[\frac{M_s}{M_c} = \frac{(\frac{1}{4})^2}{\pi (\frac{1}{2\pi})^2}\] After calculating the ratio, we arrive at the following: \[\frac{M_s}{M_c} = \frac{1}{16} \cdot \frac{4\pi^2}{\pi}\] This further simplifies to: \[\frac{M_s}{M_c} = \frac{\pi}{2}\] So the ratio of the magnetic moment of the square to the circle is \(\pi: 2\), which corresponds to option (b).

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Most popular questions from this chapter

A long wire carr1es a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is \(\mathrm{B}\). It is then bent into a circular Loop of n turns. The magnetic field at the centre of the coil for same current will be. (a) \(\mathrm{nB}\) (b) \(\mathrm{n}^{2} \mathrm{~B}\) (c) \(2 \mathrm{nB}\) (d) \(2 \mathrm{n}^{2} \mathrm{~B}\)

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) $\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)$

A Galvanometer of resistance \(15 \Omega\) is connected to a battery of 3 volt along with a resistance of \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) \(6050 \Omega\) (b) \(4450 \Omega\) (c) \(5050 \Omega\) (d) \(5550 \Omega\)

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)
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