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Chapter 13: Problem 1902

A straight wire of length \(30 \mathrm{~cm}\) and mass 60 milligram lies in a direction \(30^{\circ}\) east of north. The earth's magnetic field at this site is horizontal and has a magnitude of \(0.8 \mathrm{G}\). What current must be passed through the wire so that it may float in air ? \(\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (a) \(10 \mathrm{Amp}\) (b) \(20 \mathrm{Amp}\) (c) \(40 \mathrm{Amp}\) (d) \(50 \mathrm{Amp}\)

Short Answer

Expert verified
The required current for the wire to float in air is approximately \(50 A\), which corresponds to option (d).

Step by step solution

01

Calculate the weight of the wire

First, we need to find the gravitational force acting on the wire. The weight of the wire can be determined using the formula: Weight (W) = mass (m) × gravitational acceleration (g) Given, mass of the wire (m) = 60 milligram = 60 × 10^(-3) kg (converting to kg) Gravitational acceleration (g) = 10 m/s² Weight (W) = 60 × 10^(-3) kg × 10 m/s² = 0.6 N
02

Set up the equilibrium condition

The wire floats in the air due to the balance between the magnetic force (F_m) and the weight of the wire (W). The magnetic force on the wire can be calculated using this formula: F_m = B × I × L × sin(θ) Where: - B is the magnetic field (0.8 G = 0.8 × 10^(-4) T, converting to Tesla) - I is the current that we need to find - L is the length of the wire (30 cm = 0.3 m, converting to meters) - θ is the angle between the wire and the direction of the magnetic field (30 degrees) Since the wire is floating, the magnetic force (F_m) is equal to the weight (W), so: F_m = W
03

Solve for the current (I)

Using the equilibrium condition (F_m = W), we can solve for the current (I): 0.8 × 10^(-4) T × I × 0.3 m × sin(30°) = 0.6 N Dividing both sides by (0.8 × 10^(-4) T × 0.3 m × sin(30°)), we get: I = 0.6 N / (0.8 × 10^(-4) T × 0.3 m × sin(30°)) I = 0.6 N / (0.00008 T × 0.3 m × 0.5) I ≈ 50 A So, the required current for the wire to float in air is approximately 50 A, which corresponds to option (d).

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Most popular questions from this chapter

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\( at an angle of \)60^{\circ}\( to the \)\mathrm{X}$ -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of \(r=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $4 \pi \times 10^{-7} \mathrm{sec}$

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

The mag. field due to a current carrying circular Loop of radius $3 \mathrm{~cm}\( at a point on the axis at a distance of \)4 \mathrm{~cm}$ from the centre is \(54 \mu \mathrm{T}\) what will be its value at the centre of the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of $40 \mathrm{~cm}$, the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)

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) \(0.3 \mathrm{~N}\)
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