Moving Charges and Magnetism

The magnetic field at a point $ P $ due to a current carrying straight wire, such that the ends of the wires, makes angle $ \alpha $ and $ \beta $ (see figure) is given by

$ B=\frac{\mu_{0} I}{4 \pi d}(\sin \alpha-\sin \beta) $

where, $ d $ is the distance between the point $ P $ and wire.

In the given problem, $ \alpha=\theta_{1} $ and $ \beta=\theta_{2} $

$ \therefore \quad B=\frac{\mu_{0} I}{4 \pi d}\left(\sin \theta_{1}-\sin \theta_{2}\right) $

Percentage change in kinetic energy $=16 \%$.

As, $\quad r=\frac{m v}{q B}=\frac{\sqrt{2 m k}}{q B}$

$\Rightarrow \quad r \propto(k)^{1 / 2} \Rightarrow \frac{\Delta r}{r} \times 100=\frac{1}{2} \frac{\Delta k}{k} \times 100$

$\Rightarrow \%$ change in radius $=\frac{1}{2} \times 16 \%=8 \%$

Force on a moving charged particle in uniform magnetic field

$F = Bqv\sin \theta $ ..... (i)

Since, charge particle moves along the axis of circular current carrying loop, therefore, $\theta$ = 0$^\circ$ or 180$^\circ$

When $\theta$ = 0$^\circ$; $F = Bqv\sin 0^\circ $

F = 0

When $\theta$ = 180$^\circ$, $F = Bqv\sin 180^\circ $

F = 0

Force on charge particle in magnetic field is balanced by centripetal force, so ${{m{v^2}} \over r} = qvB\sin 90^\circ $

[$\theta$ = 90$^\circ$, therefore magnetic field is perpendicular]

${{m{v^2}} \over r} = qvB$ [sin90$^\circ$ = 1]

$ \Rightarrow r = {{mv} \over {qB}}$

For both particles, q and B are same.

Therefore, r $\propto$ mv

or, ${{{r_A}} \over {{r_B}}} = {{{m_A}{v_A}} \over {{m_B}{v_B}}}$

As, r_A > r_B [given]

$ \Rightarrow {m_A}{v_A} > {m_B}{v_B}$

Given, current in the wire, I = 10A

Radius of the wire, R = 20$\pi$ cm = 20$\pi$ $\times$ 10^$-$2 m

Magnetic field produced by semi-circular current carrying thin wire at centre O,

$B = {{{\mu _0}I} \over {4R}}$

Substituting all the given values in above equation, we get

$B = {{4\pi \times {{10}^{ - 7}} \times 10} \over {4 \times 20\pi \times {{10}^{ - 2}}}}$

B = 5 $\times$ 10^$-$6 T

Now, magnitude of force per unit length,

F = IB = 10 $\times$ 5 $\times$ 10^$-$6

= 50 $\times$ 10^$-$6 N/m

= 50 $\mu$N/m