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Answers without the blur. Sign up and see all textbooks for free! Q. 55

Expert-verified Found in: Page 1 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # A $20$-cm-long, zero-resistance slide wire moves outward, on zero-resistance rails, at a steady speed of $10m/s$ in a $0.10T$ magnetic field. On the opposite side, a $1.0Ω$ carbon resistor completes the circuit by connecting the two rails. The mass of the resistor is $50mg$. a. What is the induced current in the circuit? b. How much force is needed to pull the wire at this speed? c. If the wire is pulled for $10s$ , what is the temperature increase of the carbon? The specific heat of carbon is $710J/kgK$

(a) Induced current, ${I}_{\text{induced}}=0.2\mathrm{A}$

(b) Force, $F=4×{10}^{-4}\mathrm{N}$

(c) Increase in temperature, $\Delta T=11\mathrm{K}$

See the step by step solution

## Step 1: Introduction (part a)

a.

The amount of magnetic field that flows through a loop of area $A$ is called the magnetic flux $\Phi$. The magnetic flux is provided by when the magnetic field makes an angle with the plane.

${\Phi }_{\mathrm{m}}=BA$

The loop moves with speed $v$ for distance $x$ in time $t$. So, the distance is $x=vt$. When the loop is complete, this distance denotes the side of the loop. So, the area of the loop will be,

$A=lx=lvt$

The induced emf is the change in magnetic flux inside the loop, as defined by Faraday's law, and it is given by an equation in the form

$\epsilon =\frac{d{\Phi }_{\mathrm{m}}}{dt}=B\frac{dA}{dt}=B\frac{d}{dt}\left(lvt\right)=Blv$

We apply Ohm's law to get the induced current $I$ via the inner loop, as indicated in the next equation.

${I}_{\text{induced}}=\frac{\epsilon }{R}=\frac{Blv}{R}$

To calculate the induced current, we plug the values for $l,v,B,andR$ into equation.

${I}_{\text{induced}}=\frac{Blv}{R}$

$=\frac{\left(0.10\mathrm{T}\right)\left(0.20\mathrm{m}\right)\left(10\mathrm{m}/\mathrm{s}\right)}{1\Omega }$

$=0.2\mathrm{A}$

## Step 2: Find Force (part b)

b.

a current-carrying wire moves through a magnetic flux, the field exerts a magnetic pull on the wire. The magnetic force exerted by a wire of length $l$ is given by equation,

${\stackrel{\to }{F}}_{\text{wire}}=I\stackrel{\to }{l}×\stackrel{\to }{B}=IlB\mathrm{sin}\alpha$

Where $\alpha$ is the angle formed by the current direction and the gravitational pull. When the current is parallel to the magnetic field, it is zero, but when the current is perpendicular to the magnetic field, it is greatest.

$F=IlB\mathrm{sin}\alpha$

$=\left(0.2\mathrm{A}\right)\left(0.20\mathrm{m}\right)\left(0.10\mathrm{T}\right)\mathrm{sin}90°$

$=4×{10}^{-4}\mathrm{N}$

## Step 3: Increase in Temperature (part c)

c.

Power is calculated by,

$P={I}_{\text{induced}}^{2}R=\left(0.2\mathrm{A}{\right)}^{2}\left(1\Omega \right)=0.04\mathrm{W}$

The rate at which energy or work is transferred is referred to as power. Because this energy dissipated as heat $Q$, the heat is calculated as follows:

$Q=Pt=\left(0.04\mathrm{W}\right)\left(10\mathrm{s}\right)=0.4\mathrm{J}$

The change in temperature is obtained from the relationship $Q=mcDeltaT$ $\Delta T=\frac{Q}{mc}$

$=\frac{0.4\mathrm{J}}{\left(50×{10}^{-6}\mathrm{kg}\right)\left(710\mathrm{J}/\mathrm{kg}/\mathrm{K}\right)}$

$=11\mathrm{K}$ ### Want to see more solutions like these? 