(a) (i) Write the principle and explain the working of a moving coil galvanometer. A galvanometer as such cannot be used to measure the current in a…

Principle – When a rectangular loop carrying current I is placed in a uniform magnetic field, it experiences a torque.

Working:-

When a current flows through the coil of a galvanometer, a torque acts on it.

$\displaystyle \tau = NiAB\sin\theta$

For radial magnetic field; $\displaystyle \sin\theta = 1$

The spring provides a counter or restoring toqrue $\displaystyle k\phi$.

\[k\phi = NiAB \]

In equilibrium; $\displaystyle \phi = \left(\dfrac{NAB}{k}\right) i$

Galvanometer cannot be used as such to measure current because:

The magnetic field is made radial in a moving coil galvanometer so that the magnetic dipole moment ($\displaystyle \vec{m}$) is always perpendicular to the magnetic field ($\displaystyle \vec{B}$) Hence, $\displaystyle \sin\theta = 1$ always

Alternatively: The magnetic field is made radial in a moving coil galvanometer to make the scale linear.

It is achieved by using curved magnetic poles.

Alternatively:- By using soft iron cylindrical core.

\[dB = \frac{\mu_0}{4\pi}\frac{I\,|d\mathbf{l}\times\mathbf{r}|}{r^{3}} \]

\[= \frac{\mu_o\, i\, dl \sin 90^{o}}{4\pi\left(x^{2}+R^{2}\right)} \]

$\displaystyle dB_{\perp}$ cancels out.

Net B $\displaystyle = \int dB_x = \int dB\cos\theta$

\[= \frac{\mu_o}{4\pi}\int \frac{i\,dl}{\left(x^{2}+R^{2}\right)} \times \frac{R}{\left(x^{2}+R^{2}\right)^{1/2}} \]

\[= \frac{\mu_o i R}{4\pi\left(x^{2}+R^{2}\right)^{3/2}} \int dl \]

\[= \frac{\mu_o i R}{4\pi\left(x^{2}+R^{2}\right)^{3/2}}\,(2\pi R) \]

\[\mathbf{B} = B_x\hat{\mathbf{i}} = \frac{\mu_0 I R^{2}}{2\left(x^{2}+R^{2}\right)^{3/2}}\hat{\mathbf{i}} \]

Differences