For rotational equilibrium of the sphere about point $$P.$$
$$\eqalign{ & \left( {mg\sin \theta } \right)R = M\left( {B\sin \theta } \right) \cr & {\text{or,}}\,\,mgR\sin \theta = i\pi {R^2}B\sin \theta \cr & B = \frac{{mg}}{{\pi iR}} \cr} $$

As the electron beam is not deflected, then $${F_m} = {F_e}\,\,{\text{or}}\,\,Bev = Ee$$
or $$v = \frac{E}{B}\,......\left( {\text{i}} \right)$$
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If $$V$$ is the potential difference between the anode and cathode, then potential energy of the electron at cathode $$= eV.$$  Also, kinetic energy of the electron at anode $$ = \frac{1}{2}m{v^2}.$$   According to law of conservation of energy.
$$\frac{1}{2}m{v^2} = eV\,\,{\text{or}}\,\,v = \sqrt {\frac{{2eV}}{m}} \,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we have
$$\sqrt {\frac{{2eV}}{m}} = \frac{E}{B}\,\,{\text{or}}\,\,\frac{e}{m} = \frac{{{E^2}}}{{2V{B^2}}}$$

For stable equilibrium $$\left. {\vec M} \right\|\vec B$$
For unstable equilibrium $$\left. {\vec M} \right\|\left( { - \vec B} \right)$$

Given : Magnetic moment, $$M = 6.7 \times {10^{ - 2}}A{m^2}$$
Magnetic field, $$B = 0.01T$$
Moment of inertia, $$I = 7.5 \times {10^{ - 6}}Kg{m^2}$$
Using, $$T = 2\pi \sqrt {\frac{I}{{MB}}} $$
$$ = 2\pi \sqrt {\frac{{7.5 \times {{10}^{ - 6}}}}{{6.7 \times {{10}^{ - 2}} \times 0.01}}} = \frac{{2\pi }}{{10}} \times 1.06\,s$$
Time taken for 10 complete oscillations
$$\eqalign{ & t = 10T = 2\pi \times 1.06 \cr & = 6.6568 \approx 6.65\,s \cr} $$

The force per unit length is given by
$$\eqalign{ & f = \frac{{{\mu _0}{i^2}}}{{2\pi d}} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,f \propto {i^2} \cr} $$

Galvanometer is converted into ammeter, by connected a shunt, in parallel with it.

$$\eqalign{ & \frac{{GS}}{{G + S}} = \frac{{{V_G}}}{I} = \frac{{25 \times {{10}^{ - 3}}}}{{25}} \cr & \frac{{GS}}{{G + S}} = 0.001\,\Omega \cr & {\text{Here}}\,S < < G\,{\text{so}}\,S = 0.001\,\Omega \cr} $$

For parallel, $$M$$ is stable and for antiparallel, it is unstable.

$$\eqalign{ & {I_g} = 0.1\,I,{I_s} = 0.9I;S = \frac{{{I_g}{R_g}}}{{{I_s}}} \cr & = \frac{{0.1 \times 900}}{{0.9}} = 100\,\Omega \cr} $$

KEY CONCEPT : Use Fleming's left hand rule. We find that a force is acting in the radially outward direction throughout the circumference of the conducting loop.

According to Biot-Savart law, the magnetic field induction $$dB$$  (also called magnetic flux density) at a point $$P$$ due to current element depends upon the factors as stated below.
(i) $$dB \propto i$$
(ii) $$dB \propto dl$$
(iii) $$dB \propto \sin \theta $$
(iv) $$dB \propto \frac{1}{{{r^2}}}$$
Combining these factors, we get magnitude of $$dB$$
i.e., $$dB = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{i\,dl\sin \theta }}{{{r^2}}}$$
In vector form,
$$dB = \frac{{{\mu _0}}}{{4\pi }}i\frac{{\left( {dl \times r} \right)}}{{{r^3}}}\,\,\,\left[ {\frac{{{\mu _0}}}{{4\pi }} = {{10}^{ - 7}}T{\text{ - }}m/A} \right]$$