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

From ampere circuital law
$$\oint {B.dl} = {\mu _0}i$$
Where, $$i$$ = Net current passing through the loop
Now $$i = I + I - 2I = 0$$
$$\therefore B.2\pi \,r = 0 \Rightarrow B = 0$$

K.E. of electron $$ = 10\,eV \Rightarrow \frac{1}{2}m{v^2} = 10\,eV$$
$$\eqalign{ & \Rightarrow \frac{1}{2}\left( {9.1 \times {{10}^{ - 31}}} \right){v^2} = 10 \times 1.6 \times {10^{ - 19}} \cr & \Rightarrow {v^2} = \frac{{2 \times 10 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}} \cr & \Rightarrow {v^2} = 3.52 \times {10^{12}} \Rightarrow v = 1.88 \times {10^6}\,m \cr} $$
Also, we know that for circular motion
$$\frac{{m{v^2}}}{r} = Bev \Rightarrow r = \frac{{mv}}{{Be}} = 11\,cm$$

The magnetic field, due the coil, carrying current $$I$$ Ampere
$${B_1} = \frac{{{\mu _0}I}}{{2R}}$$
The magnetic field due to the coil, carrying current $$2\,I$$  Ampere
$${B_2} = \frac{{{\mu _0}\left( {2I} \right)}}{{2R}}$$
The resultant $$B$$
$$\eqalign{ & {B_{{\text{net}}}} = \sqrt {B_1^2 + B_2^2 + 2{B_1}{B_2}\cos \theta ,} \,\theta = {90^ \circ } \cr & {B_{{\text{net}}}} = \sqrt {B_1^2 + B_2^2} = \frac{{{\mu _0}\left( {2I} \right)}}{{2R}}\sqrt {1 + 4} \cr & = \frac{{\sqrt 5 {\mu _0}I}}{{2R}} \cr} $$

$$\eqalign{ & \frac{{{B_2}}}{{{B_1}}} = \frac{{{\mu _0}{n_2}{i_2}}}{{{\mu _0}{n_1}{i_1}}} \Rightarrow \frac{{{B_2}}}{{6.28 \times {{10}^{ - 2}}}} = \frac{{100 \times \frac{i}{3}}}{{200 \times i}} \cr & {B_2} = \frac{{6.28 \times {{10}^{ - 2}}}}{6} = 1.05 \times {10^{ - 2}}Wb/{m^2} \cr} $$

When a test charge $${q_0}$$ enters a magnetic field $${\vec B}$$ directed along $$z$$-axis, with a velocity $${\vec v}$$ making angles $$d$$ with the $$z$$-axis. The time period of the motion is independent of $$R$$ and $$v.$$

$$T = 2\pi \sqrt {\frac{I}{{M \times B}}} \,{\text{where}}\,I = \frac{1}{{12}}m{\ell ^2}$$
When the magnet is cut into three pieces the pole strength will remain the same and
M.I. $$\left( {I'} \right) = \frac{1}{{12}}\left( {\frac{m}{3}} \right){\left( {\frac{\ell }{3}} \right)^2} \times 3 = \frac{I}{9}$$
We have, Magnetic moment $$\left( M \right)$$ = Pole strength $$\left( m \right) \times \ell $$
∴ New magnetic moment,
$$M' = m \times \left( {\frac{\ell }{3}} \right) \times 3 = m\ell = M\,\,\,\therefore T' = \frac{T}{{\sqrt 9 }} = \frac{2}{3}s.$$

$$\eqalign{ & F = {F_1} - {F_2} = \frac{{{\mu _0}}}{{2\pi }}\left( {\frac{{{I_1}{I_2}}}{a} - \frac{{{I_1}{I_2}}}{{2a}}} \right) \times a \cr & = \frac{{{\mu _0}{i_1}{i_2}}}{{4\pi }}\left( {{\text{Repulsive}}} \right) \cr} $$

Corecivity, $$H = \frac{B}{{{m_0}}}\,{\text{and}}\,B = {\mu _0}ni\left( {n = \frac{N}{\ell }} \right)$$
$${\text{or,}}\,H = \frac{N}{\ell }i = \frac{{100}}{{0.2}} \times 5.2 = 2600\,A/m$$

Magnetic induction at the centre of current carrying coil of $$N$$ turns carrying current is given by
$$B = \frac{{{\mu _0}Ni}}{{2r}}\,\,......\left( {\text{i}} \right)$$
Suppose the length of the wire be $$L.$$
I
For coil of one turn, let radius be $${r_1}.$$
$$\eqalign{ & \therefore L = 2\pi {r_1} \times N \cr & {\text{or}}\,\,{r_1} = \frac{L}{{2\pi \times N}} = \frac{L}{{2\pi }}\,\,\left( {\because N = 1} \right) \cr} $$
II
For coil of two turns, let radius be $${r_2}.$$
$$\eqalign{ & \therefore L = 2\pi {r_2} \times N \cr & {\text{or}}\,\,{r_2} = \frac{L}{{2\pi \times N}} = \frac{L}{{2\pi \times 2}}\,\,\left( {\because N = 2} \right) \cr & {\text{or}}\,\,{r_2} = \frac{{{r_1}}}{2} \cr} $$
By comparing two different cases from Eq. (i),
$$\eqalign{ & \frac{{{B_1}}}{{{B_2}}} = \frac{{{N_1}}}{{{r_1}}} \times \frac{{{r_2}}}{{{N_2}}}\,\,{\text{or}}\,\,\frac{{{B_1}}}{{{B_2}}} = \frac{{1 \times \frac{{{r_1}}}{2}}}{{{r_1} \times 2}} \cr & \therefore \frac{{{B_1}}}{{{B_2}}} = \frac{1}{4} \cr} $$

Given :
Magnetic field $$B = 2 \times {10^{ - 4}}weber/{m^2}$$
Velocity of electron, $$v = {10^7}\,m/s$$
Lorentz force $$F = qvB\sin \theta $$
$$\eqalign{ & = 1.6 \times {10^{ - 19}} \times {10^7} \times 2 \times {10^{ - 4}}\,\left( {\because \theta = {{90}^ \circ }} \right) \cr & = 3.2 \times {10^{ - 16}}N \cr} $$