91影视

(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.

${\mathrm{桅}}_{\mathrm{m}}=BA$

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

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

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

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

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

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

$=0.2\mathrm{A}$

(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

${\stackrel{鈫�}{F}}_{\text{wire}}=I\stackrel{鈫�}{l}脳\stackrel{鈫�}{B}=IlB\sin 鈦�伪$

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

$F=IlB\sin 鈦�伪$

$\begin{array}{r}=\left(0.2\mathrm{A}\right)\left(0.20\mathrm{m}\right)\left(0.10\mathrm{T}\right)\sin 鈦�{90}^{鈭�}\\ =4脳{10}^{鈭�4}\mathrm{N}\end{array}$

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

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

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 $Q=mc螖T$, we get the change in the temperature by

$\mathrm{螖}T=\frac{Q}{mc}$

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