Question

I A 20-cm-long, zero-resistance slide wire moves outward, on zero-resistance rails, at a steady speed of $10\mathrm{m}/\mathrm{s}$in a $0.10\mathrm{T}$magnetic field. (See Figure 30.26.) On the opposite side, a $1.0\Omega$carbon resistor completes the circuit by connecting the two rails. The mass of the resistor is $50\mathrm{mg}$.

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 $10\mathrm{s}$, what is the temperature increase of the carbon? The specific heat of carbon is$710\mathrm{J}/\mathrm{kgK}$.

Short answer

(a)The induced current in the circuit$I_{\text{induced}}=0.2\mathrm{A}$

(b) The speed is$F=4\times {10}^{-4}\mathrm{N}\text{.}$

(c)The temperature$\mathrm{\Delta }T=11\mathrm{K}$

Step-by-step solution

Find induced current in the circuit(part a)

(a) The magnetic flux is the amount of magnetic field that flows through a loop of area A. The magnetic flux is given when the magnetic field forms an angle with the plane.

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

The loop moves at speed v for a distance x in time t. So the distance is. This distance represents the side of the loop when the a whole loop is inserted into to the magnetic field. Like a result, 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 equation (30.14) in the form

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

$I_{\text{induced}}=\frac{\epsilon }{R}=\frac{Blv}{R}$

We plug the values for l, v, B and R into equation (2) to get the induced current by

$I_{\text{induced}}=\frac{Blv}{R}$

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

$=0.2\mathrm{A}$

Speed of fire(part b)

(b) When a wire carrying a current and moves inside a magnetic field, the magnetic field exerted a magnetic force on the wire. For a wire with length l, the exerted magnetic force is given by equation (29.26) in the form

${\overrightarrow{F}}_{\text{wire}}=I\overrightarrow{l}\times \overrightarrow{B}=IlB\sin \alpha$

The angle is $\alpha =90°$, so we plug the values for I, l, B and $\alpha$into equation (3) to get F by

$F=IlB\sin \alpha$

$\begin{array}{r}=\left(0.2\mathrm{A}\right)\left(0.20\mathrm{m}\right)\left(0.10\mathrm{T}\right)\sin {90}^{\circ }\\ =4\times {10}^{-4}\mathrm{N}\end{array}$

Step 3:Find temputure increase of the carbon(part c)

(c) The carbon rod's resistance dissipated a power P, which is calculated as

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

Power is defined as the rate at which energy or work is transferred. This energy was dissipated as heat Q, and this heat is calculated as

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

The heat increases the temperature, from the relation $Q=mc\Delta T$, we get the change in the temperature by

$\mathrm{\Delta }T=\frac{Q}{mc}$

$\begin{array}{r}=\frac{0.4\mathrm{J}}{\left(50\times {10}^{-6}\mathrm{kg}\right)(710\mathrm{J}/\mathrm{kg}/\mathrm{K})}\\ =11\mathrm{K}\end{array}$