/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A straight wire of mass \(200 \m... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 47

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5 \mathrm{~m}\) carries a current of \(2 \mathrm{~A}\). It is suspended in mid-air by a uniform horizontal magnetic field \(B\). The magnitude of \(B\) (in tesla) is $$ \left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right) $$ (A) 2 (B) \(1.5\) (C) \(0.55\) (D) \(0.66\)

Short Answer

Expert verified
The magnitude of the magnetic field needed to keep the wire suspended in mid-air is approximately 0.65 T which is closest to the answer choice (D) 0.66.

Step by step solution

01

Calculate the gravitational force on the wire

First, we need to find the gravitational force acting on the wire. The force of gravity (F_gravity) can be calculated using the formula: $$ F_\text{gravity} = m \times g $$ where m is the mass of the wire (in kg) and g is the acceleration due to gravity (in m/s^2). The mass of the wire is given as 200 gm, so we need to convert it to kg: $$ m = 200~\text{gm} \times \frac{1~\text{kg}}{1000~\text{gm}} = 0.2~\text{kg} $$ The acceleration due to gravity, g, is given as 9.8 m/s^2. Plugging these values into the formula, we can find the gravitational force: $$ F_\text{gravity} = 0.2~\text{kg} \times 9.8 \frac{\text{m}}{\text{s}^2} = 1.96 \text{N} $$
02

Equate the magnetic force to the gravitational force

The force on a current-carrying wire in a magnetic field (F_magnetic) can be calculated using the formula: $$ F_\text{magnetic} = B \times I \times L $$ where B is the magnetic field (in Tesla), I is the current (in Ampere), and L is the length of the wire (in meters). We know that the magnetic force has to be equal to the gravitational force in order to keep the wire suspended in mid-air: $$ F_\text{magnetic} = F_\text{gravity} $$ which means: $$ B \times I \times L = m \times g $$ We have already found m 脳 g, which is equal to 1.96 N, and we are given the length of the wire, L = 1.5 m, and the current, I = 2 A.
03

Solve for the magnetic field B

Now, we can solve the equation for the magnetic field, B: $$ B = \frac{m \times g}{I \times L} $$ Plugging in the values: $$ B = \frac{1.96~\text{N}}{2~\text{A} \times 1.5~\text{m}} $$ $$ B = \frac{1.96}{3} ~\text{T} $$ $$ B 鈮� 0.65 ~\text{T} $$ Thus, the magnitude of the magnetic field needed to keep the wire suspended in mid-air is approximately 0.65 T. The closest answer choice is (D) 0.66.

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Most popular questions from this chapter

Two circular coils of radii \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) carry equal currents of \(2 \mathrm{~A}\). The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes and their centres coincide. Magnitude of magnetic field at the common centre of coils is, (A) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in same direction. (B) \(4 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction. (C) zero, if currents in the coils are in opposite direction. (D) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction.

A conductor \(A B\) carries a current \(i\) in a magnetic field \(\vec{B}\). If \(\overrightarrow{A B}=\vec{r}\) and the force on the conductor is \(\vec{F}\) (A) \(\vec{F}\) does not depend on the shape of \(A B\). (B) \(\vec{F}=i(\vec{r} \times \vec{B})\). (C) \(\vec{F}=i(\vec{B} \times \vec{r})\). (D) \(|\vec{F}|=i(\vec{r} \cdot \vec{B})\).

Through two parallel wires \(A\) and \(B, 10\) and \(2 \mathrm{~A}\) of currents are passed, respectively, in opposite direction. If the wire \(A\) is infinitely long and the length of the wire \(B\) is \(2 \mathrm{~m}\), the force on the wire \(B\), which is situated at \(10 \mathrm{~cm}\) distance from \(A\) will be (A) \(8 \times 10^{-5} \mathrm{~N}\) (B) \(4 \times 10^{-7} \mathrm{~N}\) (C) \(4 \times 10^{-5} \mathrm{~N}\) (D) \(4 \pi \times 10^{-7} \mathrm{~N}\)

Two short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\) such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \frac{M}{2 d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \frac{M^{2}}{2 d^{3}}\)

Relative permittivity and permeability of a material \(\varepsilon_{r}\) and \(\mu_{r}\) respectively. Which of the following value of these quantities are allowed for a diamagnetic material? (A) \(\varepsilon_{r}=0.5, \mu_{r}=1.5\) (B) \(\varepsilon_{r}=1.5, \mu_{r}=0.5\) (C) \(\varepsilon_{r}=0.5, \mu_{r}=0.5\) (D) \(\varepsilon_{r}=1.5, \mu_{r}=1.5\)
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