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:

Answer 1

Step 1: Given data

Number 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

$$\begin{gathered} \Rightarrow i_0=\frac{E_0}{R}[where{i}_0=peakcurrent,E_0=inducedemf,R=resis\tan ce] \\ \Rightarrow E_0=NB{A}_{}[where{E}_0=inducedemf,N=numberofturns,B-magneticflux,A=area] \\ \Rightarrow R_{total}=R+R\text{'}[R=resis\tan ce] \end{gathered}$$

Step 3: Calculating induced emf

The values given are, $N=500,A=50c{m}^2=50\times {10}^{-4m^2},B=0.14Wb/m^2,\omega =150rad/s$

Substitute it in the induced emf formula

$$\begin{gathered} E_0=NBA \\ {E}_0=500\times 50\times {10}^{-4}\times 0.14\times 150 \\ {E}_0=52.5V \end{gathered}$$

Step 4: Calculating total resistance

$$\begin{gathered} R_{total}=R+R\text{'} \\ {R}_{total}=5\Omega +10\Omega \\ {R}_{total}=15\Omega \end{gathered}$$

Step 4: Calculating peak current

$$\begin{gathered} i_0=\frac{E_0}{R} \\ {i}_0=\frac{52.5}{15} \\ {i}_0=3.5A \end{gathered}$$

Hence, the value of peak current through the resistance is $3.5A$.