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Chapter 13: Problem 1885

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

Short Answer

Expert verified
The magnetic field B required to suspend the wire in mid-air is (a) \(\frac{2}{3} \text{ tesla}\).

Step by step solution

01

Convert the mass of the wire to kg

Since the mass of the wire is given in grams, we need to convert it to kg for our calculations. To do so, divide the mass by 1000. \(m = \frac{200}{1000} kg = 0.2 kg\)
02

Calculate the gravitational force acting on the wire

The gravitational force acting on the wire (its weight) can be found using the formula: \(F_g = m \cdot g\) where m is the mass of the wire and g is the acceleration due to gravity (given as 10 m/s²). \(F_g = (0.2 kg)(10 \frac{m}{s^2}) = 2 N\)
03

Write down the formula for the magnetic force on the wire

The magnetic force F_B acting on a current-carrying wire is given by the formula: \(F_B = B \cdot I \cdot L\) where B is the magnetic field, I is the current passing through the wire, and L is the length of the wire.
04

Equate the magnetic force to the gravitational force

Since the magnetic force is suspending the wire in mid-air, it must be equal to the gravitational force acting on the wire: \(F_B = F_g\)
05

Solve for the magnetic field B

Substitute the known values in the equation: \(B \cdot I \cdot L = F_g\) \(B \cdot (2 A) \cdot (1.5 m) = 2 N\) Now, solve for B: \(B = \frac{2 N}{(2 A)(1.5 m)} = \frac{2}{3} T\) So the magnetic field B required to suspend the wire in mid-air is (a) \(\frac{2}{3} \text{ tesla}\).

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Most popular questions from this chapter

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