Question

It is desired to measure the magnitude of field between the poles of a powerful loudspeaker magnet. A small flat search coil of area $2{\text{cm}}^2$ with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick ${90}^{\circ }$ turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is $7.5mC$. The combined resistance of the coil and the galvanometer is $0.50\Omega$. Estimate the field strength of magnet.

Answer 1

Step 1: Find induced current in the coil.
According to Faraday’s law, Induced emf is given by
$ϵ=-N\frac{d\varphi }{dt}$
Where,
$\frac{d\varphi }{dt}$ = Rate of change in flux
Induced current in the coil.
$I=\frac{ϵ}{R}$
Combining equations for $ϵ$ and I, we get,
$I=-\frac{N}{R}\frac{d\varphi }{dt}$
Step 2: Find charge flown in the coil.
From the expression for induced current,
$Idt=-\frac{N}{R}d\varphi$
Integrating the above equation, we get,
$\int Idt=-\frac{N}{R}{\int }_{{\varphi }_1}^{{\varphi }_2}d\varphi$
$\Rightarrow It=\frac{N}{R}({\varphi }_1-{\varphi }_2)$
$\Rightarrow Q=\frac{N}{R}({\varphi }_1-{\varphi }_2)$
Step 3: Calculate the value of magnetic field strength.
Now, initial flux
through the coil,
${\varphi }_1=BA$
And final flux through the coil, ${\varphi }_2=0$ (because angle between magnetic field and area vector is ${90}^{\circ })$
$\Rightarrow Q=\frac{NBA}{R}$
$\Rightarrow B=\frac{QR}{NA}$
Given, area of the small flat search coil,
A = 2 ${\text{cm}}^2=2\times {10}^{-4}{m}^2$
Number of turns on the coil, $N=25$
Total charge flowing in the coil, $Q=7.5mC=7.5\times {10}^{-3}C$
Total resistance of the coil and galvanometer, $R=0.50\Omega$
Putting the values, we get,
$\Rightarrow B=\frac{7.5\times {10}^{-3}\times 0.50}{25\times 2\times {10}^{-4}}=0.75T$
Therefore, the field strength of the magnet is $0.75T$.
Final answer: $0.75T$