As revolving charge is equivalent to a current, so $$ = qf = q \times \frac{\omega }{{2\pi }}$$
But $$\omega = \frac{v}{R}$$
where, $$R$$ is radius of circle and $$v$$ is uniform speed of charged particle.
Therefore, $$i = \frac{{qv}}{{2\pi R}}$$
Now, magnetic moment associated with charged particle is given by
$$\mu = iA = i \times \pi {R^2}\,\,{\text{or}}\,\,\mu = \frac{{qv}}{{2\pi R}} \times \pi {R^2} = \frac{1}{2}qvR$$
Alternative
Current produced due to circular motion of charge $$q$$ is given by
$$i = \frac{q}{T}\,\,\left[ {T = {\text{Time period}}\,{\text{of revolution}}} \right]$$
Now, $$T = \frac{{2\pi R}}{v}$$
So, $$i = \frac{q}{{2\pi R}} \times v$$
Now, magnetic moment $$\left( \mu \right)$$ is given by
$$\mu = iA$$
So, $$\mu = \frac{{qv}}{{2\pi R}} \times \pi {R^2}$$
$$ \Rightarrow \mu = \frac{{qvR}}{2}$$

Time period,
$$T = \frac{{2\pi m}}{{qB}} = \frac{{2\pi }}{{\pi \times 2}} = 1\,s$$
Thus, particle will be at point $$P$$ after $$t = \frac{1}{{12}}s$$
$$\eqalign{ & \vec v = 10\left[ {\cos 30\hat i + \sin 30\hat j} \right] \cr & \vec v = 10\left[ {\frac{{\sqrt 3 }}{2}\hat i + \frac{1}{2}\hat j} \right] = 5\left[ {\sqrt 3 \hat i + \hat j} \right]m{s^{ - 1}} \cr} $$

$$\eqalign{ & \vec B = - c\hat k,\,\,{\text{and}}\,\,\vec v = v\cos {45^ \circ }\hat i - v\sin {45^ \circ }\hat j \cr & = \frac{v}{{\sqrt 2 }}\hat i - \frac{v}{{\sqrt 2 }}\hat j. \cr & {\text{Thus,}}\,\,\vec F = q\left( {\vec v \times \vec B} \right) = q\left[ {\left( {\frac{v}{{\sqrt 2 }}\hat i - \frac{v}{{\sqrt 2 }}\hat j} \right) \times \left( { - c\hat k} \right)} \right] \cr & = qcv\left[ {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right] \cr & \therefore \vec a = \left[ {\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right]. \cr} $$

$$V = {i_g}\left( {G + R} \right)\,{\text{and}}\,V' = {i_g}\left( {G + 2R} \right)$$
It shows, $$V < V' < 2\,V$$

The magnetic field from the centre of wire of radius $$R$$ is given by
$$\eqalign{ & B = \left( {\frac{{{\mu _0}I}}{{2{R^2}}}} \right)r\,\,\left( {r < R} \right) \cr & \Rightarrow B \propto r \cr & {\text{and}}\,B = \frac{{{\mu _0}I}}{{2{R^2}}}\,\,\left( {r > R} \right) \cr & \Rightarrow B \propto \frac{1}{r} \cr} $$
From the above descriptions, we can say that the graph (B) is a correct representation.

$$\eqalign{ & F = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{r} \times \ell \cr & = \frac{{{{10}^{ - 7}} \times 2 \times 10 \times 2}}{{0.1}} \times 2 \cr & = 8 \times {10^{ - 5}}N \cr} $$

Magnetic field at the centre of the shell $$ = 2\frac{{{\mu _0}{i_1}}}{{{\pi ^2}r}}$$
Force per unit length of the wire $$F = B{i_2} \times 1$$
$$ = \left[ {2\frac{{{\mu _0}{i_1}{i_2}}}{{{\pi ^2}r}}} \right]$$

$$r = \frac{{\sqrt {2mv} }}{{qB}} \Rightarrow r \times v\frac{{\sqrt m }}{q}$$
Thus we have, $${r_\alpha } = {r_p} < {r_d}$$

KEY CONCEPT: $$r \propto \frac{{\sqrt m }}{q}$$
$$\eqalign{ & \therefore {r_p}:{r_d}:{r_\alpha } = \frac{{\sqrt 1 }}{1}:\frac{{\sqrt 2 }}{1}:\frac{{\sqrt 4 }}{1} = 1:\sqrt 2 :1 \cr & \Rightarrow {r_\alpha } = {r_p} < {r_d} \cr} $$

Force on a current carrying conductor placed in a magnetic field is given by
$$F = i\left( {l \times B} \right) = il\,B\sin \theta $$
where, $$\theta =$$  angle between current elements and magnetic field.
If linear conductor carrying current is placed perpendicular to the direction of magnetic field, $$\left( {\theta = {{90}^ \circ }} \right)$$   it will experience maximum force.
i.e., $${F_{\max }} = ilB$$
Given, $$i = 1.2\,A,\,l = 0.5\,m\,\,{\text{and}}\,B = 2T$$
$$\therefore F = 2 \times 1.2 \times 0.5 = 1.2\,N$$