Magnetic Effect of Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics

When charged particle move on circular path, the force $$F$$ on the charged particle due to magnetic field provides the required centripetal force $$\left( { = \frac{{m{v^2}}}{r}} \right)$$   necessary for motion along the circular path.
So, $$\frac{{m{v^2}}}{r} = qvB$$
where, $$m =$$  mass of particle
$$v =$$  velocity of particle
$$q =$$  charge on particle
$$B =$$  external magnetic field
$$r =$$  radius of circular path
$$m{v^2} = Bqvr$$
$$\therefore $$ Kinetic energy $${E_K} = \frac{1}{2}m{v^2} = \frac{1}{2}Bqvr$$
$$ = Bq \cdot \frac{r}{2}\frac{{Bqr}}{m} = \frac{{{B^2}{q^2}{r^2}}}{{2\,m}}$$
For deutron, $${E_1} = \frac{{{B^2}{q^2}{r^2}}}{{2 \times \left( {2\,m} \right)}}\,\,\left( {{\text{mass}} = 2\,m} \right)$$
For proton, $${E_2} = \frac{{{B^2}{q^2}{r^2}}}{{2\,m}}\,\,\left( {{\text{mass}} = m} \right)$$
$$\therefore \frac{{{E_1}}}{{{E_2}}} = \frac{1}{2} \Rightarrow \frac{{50\,keV}}{{{E_2}}} = \frac{1}{2}\,{\text{or}}\,\,{E_2} = 100\,keV$$

$$SI$$ unit of magnetic induction is tesla $$\left( T \right).$$
Magnetic field induction at a point is said to be one tesla if a charge of one coulomb while moving at right angle to a magnetic field, with a velocity of $$1\,m{s^{ - 1}}$$  experiences a force of $$1\,N,$$  at that point.
$$\therefore 1\,T = 1\,N{A^{ - 1}}{m^{ - 1}}$$

A magnetic needle kept in non uniform magnetic field experience a force and torque due to unequal forces acting on poles.

There will be no magnetic field at $$O$$ due to wire
$$PQ$$  and $$RS$$  Magnetic field at $$'O'$$ due
to arc $$QR = \frac{{{m_0}}}{{4p}}\frac{{\left( {10} \right)}}{{\left( {3'{{10}^{ - 2}}} \right)}}'\frac{p}{4}\,\,\,\left( {{\text{Perpendicular outwards}}} \right)$$

Magnetic field at $$'O'$$ due to arc $$PS$$
$$ = \frac{{{m_0}}}{{4p}}'\frac{{\left( {10} \right)}}{{\left( {5'{{10}^{ - 2}}} \right)}}'\frac{p}{4}\,\,\,\left( {{\text{Perpendicular inwards}}} \right)$$
∴ Net magnetic field at $$'O'$$
$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }} \times 10\left[ {\frac{1}{{\left( {3 \times {{10}^{ - 2}}} \right)}} - \frac{1}{{\left( {5 \times {{10}^{ - 2}}} \right)}}} \right] \times \frac{\pi }{4} \cr & \Rightarrow B = \frac{\pi }{3} \times {10^{ - 5}}T \approx 1 \times {10^{ - 5}}T\,\,\,\,\left( {{\text{Perpendicular outwards}}} \right) \cr} $$

$$\eqalign{ & {B_A} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{R} \times 2\pi = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{\frac{\ell }{{2\pi }}}} \times 2\pi \cr & = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{\ell } \times {\left( {2\pi } \right)^2} \cr} $$

Case (b) :
$$\eqalign{ & {B_B} = 4 \times \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{\frac{a}{2}}}\left[ {\sin {{45}^ \circ } + \sin {{45}^ \circ }} \right] \cr & = 4 \times \frac{{{\mu _0}}}{{4\pi }} \times \frac{I}{{\frac{\ell }{8}}} \times \frac{2}{{\sqrt 2 }} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{\ell } \times \root {32} \of 2 \,\,\,\,\,\left[ {4a = 1} \right] \cr} $$

KEY CONCEPT : Magentic field at the centre of a circular coil of radius $$R$$ carrying current $$i$$ is $$B = \frac{{{\mu _0}i}}{{2R}}$$
Given $$n \times \left( {2\pi r'} \right) = 2\pi r$$
$$\eqalign{ & \Rightarrow nr' = R\,......\left( 1 \right) \cr & B' = \frac{{n.{\mu _0}i}}{{2r'}}\,......\left( 2 \right) \cr} $$
from (1) and(2),
$$B' = \frac{{n{\mu _0}i.n}}{{2\pi R}} = {n^2}B$$

$$\eqalign{ & B = \frac{{{\mu _0}I}}{{4\pi r}}\left[ {2\sin \frac{\pi }{n}} \right] \cr & \cos \frac{\pi }{n} = \frac{r}{R} \cr & {\text{But}}\,{\text{or}}\,\,r = R\cos \frac{\pi }{n} \cr & \therefore B = \frac{{{\mu _0}I}}{{4\pi R\cos \frac{\pi }{n}}}\left[ {2\sin \frac{\pi }{n}} \right] = \frac{{{\mu _0}I}}{{2\pi R}}\left[ {\tan \frac{\pi }{n}} \right] \cr} $$

$$U = - \overrightarrow M .\overrightarrow B = - MB\cos \theta $$
In case I, $$\theta = {180^ \circ },U = + MB$$
In case II, $$\theta = {90^ \circ },U = 0$$
In case III, $$\theta = {\text{ acute, }}U = + ve{\text{ }}\left( {{\text{less than }} + MB} \right)$$
In case IV, $$\theta = {\text{ obtuse, }}U = - ve$$
$$\therefore I > III > II > IV.$$

$$\eqalign{ & v = \sqrt {2gR\sin \theta } \cr & {\text{and}}\,F = qvB \cr & {\text{Now}}\,N - \left( {F + mg\sin \theta } \right) = \frac{{m{v^2}}}{R} \cr & \therefore N = F + mg\sin \theta + \frac{{m{v^2}}}{R} = qB\sqrt {2gR\sin \theta } + mg\sin \theta + 2\,mg\sin \theta \cr & \therefore N = 3\,mg\sin \theta + qB\sqrt {2gR\sin \theta } \cr} $$

$$\eqalign{ & B = \frac{{{\mu _0}I}}{{4\pi d}}\left[ {\sin {\theta _1} + \sin {\theta _2}} \right] \cr & {\text{But}}\,\,{\theta _1} + {\phi _1} = {90^ \circ }\,\,{\text{or}}\,\,{\theta _1} = {90^ \circ } - {\phi _1} \cr & \sin {\theta _1} = \sin \left( {{{90}^ \circ } - {\phi _1}} \right) = \cos {\phi _1} \cr & \sin {\theta _2} = \cos {\phi _2} \cr & {B_{{\text{net}}}} = \frac{{{\mu _0}I}}{{2\pi d}}\left( {\cos {\phi _1} + \cos {\phi _2}} \right) \cr} $$