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Chapter 13: Problem 1884

A conducting rod of 1 meter length and \(1 \mathrm{~kg}\) mass is suspended by two vertical wires through its ends. An external magnetic field of 2 Tesla is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$ (a) \(0.5 \mathrm{Amp}\) (b) \(15 \mathrm{Amp}\) (c) \(5 \mathrm{Amp}\) (d) \(1.5 \mathrm{Amp}\)

Short Answer

Expert verified
The current needed to pass through the conducting rod to make the tension in the wires zero is (c) \(5 \mathrm{Amp}\).

Step by step solution

01

Analyze forces acting on the rod

First, let's consider the forces acting on the conducting rod. There are three main forces at play here: 1. Gravitational force (F_g) acting downward due to the mass of the rod 2. Magnetic force (F_m) acting upward, due to the interaction between the current and the magnetic field 3. Tension (T) in the wires supporting the rod at each end. Given that our goal is to make the tension in the wires zero, the magnetic force will have to balance the gravitational force.
02

Calculate gravitational force

To determine the gravitational force acting on the rod, use the equation: \(F_g = mg \) where m = mass of the rod and g = acceleration due to gravity. In our problem, \(m = 1 kg\) and \(g = 10 m/s^2\). Calculating the gravitational force: \(F_g = (1 kg)(10 m/s^2) = 10 N\)
03

Determine magnetic force

To find the magnetic force acting on the rod, we will use the equation: \[F_m = BIL\] where B = magnetic field strength, I = current flowing through the rod, and L = length of the rod. In our problem, the magnetic field strength is \(B = 2 T\) and the length of the rod is \(L = 1 m\). We want to find the current I that will make \(F_m = F_g\).
04

Set up the equation and solve for current I

Now, to make the tension in the wires zero, we want the magnetic force to balance the gravitational force: \[F_g = F_m\] \[mg = BIL\] Plug in the values for m, g, B, and L: \[10 N = (2 T)(1 m)(I)\] Divide both sides of the equation by \(2 T\): \[I = \frac{10 N}{2 T}\] Finally, calculate the current needed to make the tension in the wires zero: \[I = \frac{10 N}{2 T} = 5 A\] Therefore, the correct answer is (c) \(5 \mathrm{Amp}\).

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Most popular questions from this chapter

The magnetism of magnet is due to (a) The spin motion of electron (b) Earth (c) Pressure inside the earth core region (d) Cosmic rays

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) \(4 \mathrm{~m}\)

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 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) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)
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