Resistive Magnestism

Electricity and Magnetism Level pending

The magnetic flux density $B$ is changing in magnitude at a constant rate $dB/dt$ .A given mass $m$ of copper,drawn into a wire of radius $a$ and formed into a circular loop of radius $r$ is placed perpendicular to the field $B$ .The induced current in the loop is $i$ .The resistivity of copper is $p$ and the density is $d$ .The value of induced current $i$ is .. CONSIDER PI AS ¥..

(m/4¥a^2r)*(dB/dt) None of the above (m/4¥ad)*(dB/dt) (m/4¥pd)*(dB/dt)

1 solution

Rishi K
Jul 6, 2016

$E$ = $d¢/dt$ , $i$ = $E/R$ = $(1/R)*d/dt(BA)$ = $(A/R)*dB/dt$ where $¥r^2$ area of the loop of radius $r$ and resistance $R$ length= $2¥r$ and area of cross section= $¥a^2$ . .. $R$ = $(pl/(¥a^2)$ further mass of the wire = $(¥a^2)(2¥r)(d)$ substituting the value of $R$ in the first eqn we get $i$ = $((¥a^2)(¥r^2)/(4¥p))*dB/dt$ ...therefore replacing the required value to get in terms of $m$ ..we get finally $i$ = $(m/4¥pd)*db/dt$ ..

Nice solution...

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$E=\dfrac{d \phi}{dt}=\dfrac{ d(AB) }{dt}=A\dfrac{dB}{dt}$

$E= \pi r^2 \dfrac{dB}{dt}$

Now resistance of the wire, $R=\dfrac{ \rho l}{A}$

$R=\dfrac{p(2 \pi r)}{\pi a^2}$

$i=\dfrac{E}{R}=(\pi ra^2) \Bigg (\dfrac{1}{2p} \Bigg ) \Bigg ( \dfrac{dB}{dt} \Bigg ) \rightarrow$ 1

Mass of the wire, $m=(\pi a^2)(2 \pi r)(d)$

$\pi ra^2=\dfrac{m}{2 \pi d} \rightarrow$ 2

Replacing 2 in 1 ,

$i= \Bigg ( \dfrac{m}{4 \pi dp} \Bigg ) \Bigg ( \dfrac{dB}{dr} \Bigg )$

Sparsh Sarode - 4 years, 11 months ago

Resistive Magnestism

1 solution

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