    Question

# A flat coil of 500 turns each of area $50c{m}^{2}$ rotates in a uniform magnetic field of $0.14Wb/{m}^{2}$ about an axis normal to the field at an angular speed of $150rad/s$. The coil has a resistance of $5\Omega$. the induced e.m.f. is applied to an external resistance of $10\Omega$, the peak current through the resistance is:

Open in App
Solution

## Step 1: Given dataNumber of turns$\left(\mathrm{N}\right)=500$;Area $\mathrm{A}=50{\mathrm{cm}}^{2}$$\mathrm{B}=0.14\mathrm{Wb}{\mathrm{m}}^{-2}$$\mathrm{\omega }=150\mathrm{rad}{\mathrm{s}}^{-1}$$\mathrm{R}=5\mathrm{\Omega }$$\mathrm{R}\text{'}=10\mathrm{\Omega }$Step 2: Formula used$⇒{i}_{0}=\frac{{E}_{0}}{R}\left[where{i}_{0}=peakcurrent,{E}_{0}=inducedemf,R=resis\mathrm{tan}ce\right]\phantom{\rule{0ex}{0ex}}⇒{E}_{0}=NB{A}_{}\left[where{E}_{0}=inducedemf,N=numberofturns,B-magneticflux,A=area\right]\phantom{\rule{0ex}{0ex}}⇒{R}_{total}=R+R\text{'}\left[R=resis\mathrm{tan}ce\right]$Step 3: Calculating induced emfThe values given are, $N=500,A=50c{m}^{2}=50×{10}^{-4{m}^{2}},B=0.14Wb/{m}^{2},\omega =150rad/s$Substitute it in the induced emf formula${E}_{0}=NBA\phantom{\rule{0ex}{0ex}}{E}_{0}=500×50×{10}^{-4}×0.14×150\phantom{\rule{0ex}{0ex}}{E}_{0}=52.5V$Step 4: Calculating total resistance${R}_{total}=R+R\text{'}\phantom{\rule{0ex}{0ex}}{R}_{total}=5\Omega +10\Omega \phantom{\rule{0ex}{0ex}}{R}_{total}=15\Omega$Step 4: Calculating peak current${i}_{0}=\frac{{E}_{0}}{R}\phantom{\rule{0ex}{0ex}}{i}_{0}=\frac{52.5}{15}\phantom{\rule{0ex}{0ex}}{i}_{0}=3.5A$Hence, the value of peak current through the resistance is $3.5A$.  Suggest Corrections  1      Similar questions  Explore more