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Chapter 13: Problem 1969

Force between two identical bar magnets whose centers are I meter apart is \(4.8 \mathrm{~N}\), when their axes are in the same line. If separation is increased to \(2 r\), the force between them is reduced to (a) \(2.4 \mathrm{~N}\) (b) \(1.2 \mathrm{~N}\) (c) \(0.6 \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\)

Short Answer

Expert verified
The short answer is: The force between the two identical bar magnets when their separation is increased to 2 meters is \(1.2 \mathrm{~N}\).

Step by step solution

01

Identify the formula for the force between two bar magnets

The force between two identical bar magnets can be represented as follows: \(F \propto \frac{m_1 m_2}{d^2}\) Where F is the force between the magnets, m_1 and m_2 are their magnetic moments, and d is the distance between them. Since the magnets are identical, we can simplify the formula as: \(F \propto \frac{m^2}{d^2}\) Here we can introduce a constant k to convert the proportionality to an equation: \(F = k\frac{m^2}{d^2}\)
02

Calculate the constant k using the initial conditions

We know that when the distance between the magnets is 1 meter, the force between them is 4.8 N. We will use this information to find the constant k: \(4.8 = k\frac{m^2}{(1)^2}\) Solving for k, \(k = 4.8m^2\)
03

Calculate the new force when the distance is increased to 2 meters

Now, we will use the equation to find the force when the distance between the magnets is increased to 2 meters: \(F_{new} = k\frac{m^2}{(2)^2}\) Substitute the value of k from Step 2: \(F_{new} = (4.8m^2)\frac{m^2}{(2)^2}\) Solve for \(F_{new}\): \(F_{new} = \frac{4.8m^4}{4}\) Simplify the expression: \(F_{new} = 1.2m^4\) In this situation, the new force F is proportional to \(1.2m^4\). Since the proportionality constant is the same as before, the new force can be represented as: \(F_{new} = 1.2(4.8)\) \(F_{new} = 1.2 \mathrm{~N}\)
04

Choose the correct answer

From the options provided: (a) 2.4 N (b) 1.2 N (c) 0.6 N (d) 0.3 N The correct answer is (b) 1.2 N, as calculated in the previous step.

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Most popular questions from this chapter

Unit of magnetic Flux density is (a) Tesla (b) [(Weber) \(/\) (meter) \(\left.^{2}\right]\) (c) [(Newton) \(/\) (Amp - meter)] (d) All of the above

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) \(q(b+a)(B / 2 m)\)

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

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

The unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to $\mathrm{N} / \mathrm{m}$ (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)
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