Question

An earth coil has $500$ turns a resistance of $1$ ohm/turn. The average area of each turn is $0.02m^2$. When the coil is suddenly rotated through ${180}^0$, the charge induced is $4\mu C$. If the magnetic field of induction is $4\times {10}^{-4}T$, the resistance of galvanometer connected in series with the coil is:

• A. $1500$ ohm
• B. $19500$ ohm
• C. $20000$ ohm
• D. $4500$ ohm

Answer 1

The correct option is A $1500$ ohm

Given : $N=500turns$

resistance of coil,$R_c=1\ast 500=500\Omega$

Area of each turn,$A=0.02m^2$

induced charge,$\Delta q=4\ast {10}^{-6}C$

magnetic field,$B=4\ast {10}^{-4}T$

${\theta }_1=0^{\circ }$ and ${\theta }_2={180}^{\circ }$

To find : resistance of galvanometer,$R_g=?$

Solution: As we know that,

em.f. induced,$\epsilon =-N\frac{\Delta \varphi }{\Delta t}$

also,$\epsilon =\frac{\Delta q}{\Delta t}R$

So,

$\frac{\Delta q}{\Delta t}R=-N\frac{\Delta \varphi }{\Delta t}$

$==>\Delta qR=-N\Delta \varphi$ $....\left(1\right)$

Now, mean area of coil$=Areaofeachcoilperturn\ast no.ofturns/totalturns$

$==>$mean area of coil$=A\ast 500/500=A$

Now, $\Delta \varphi =finalflux-initialflux$

$==>\Delta \varphi =BAcos{\theta }_2-BAcos{\theta }_1$

$==>\Delta \varphi =BAcos180-BAcos0$

$==>\Delta \varphi =-BA-BA$

$==>\Delta \varphi =-2BA$ $....\left(2\right)$

Equating both $\left(1\right),\left(2\right)$,we get

$==>\Delta qR=2NBA$

here R is resistane and has value $R=R_c+R_g$

$==>4\ast {10}^{-6}R=2\ast 500\ast 4\ast {10}^{-4}\ast 0.02$

$==>R_c+R_g=2000$

$==>R_g=(2000-500)\Omega$

$==>R_g=1500\Omega$

hence,

The correct opt : A