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

Magnetic field at the centre of a circular coil is $$B = \frac{{{\mu _0}}}{{2\pi }} \times \frac{i}{r}$$
where, $$i$$ is current flowing in the coil and $$r$$ is radius of coil.
At the centre of coil-1,
$${B_1} = \frac{{{\mu _0}}}{{2\pi }} \times \frac{{{i_1}}}{{{r_1}}}\,......\left( {\text{i}} \right)$$
At the centre of coil-2
$${B_2} = \frac{{{\mu _0}}}{{2\pi }} \times \frac{{{i_2}}}{{{r_2}}}\,......\left( {{\text{ii}}} \right)$$
$$\eqalign{ & {\text{but}}\,\,{B_1} = {B_2} \cr & \therefore \frac{{{\mu _0}}}{{2\pi }}\frac{{{i_1}}}{{{r_1}}} = \frac{{{\mu _0}}}{{2\pi }}\frac{{{i_2}}}{{{r_2}}} \cr & {\text{or}}\,\,\frac{{{i_1}}}{{{r_1}}} = \frac{{{i_2}}}{{{r_2}}} \cr & {\text{As}}\,{\text{given}}\,\,{r_1} = 2{r_2} \cr & \therefore \quad \frac{{{i_1}}}{{2{r_2}}} = \frac{{{i_2}}}{{{r_2}}}\,\,{\text{or}}\,\,{i_1} = 2{i_2}\,......\left( {{\text{iii}}} \right) \cr} $$
Now, ratio of potential differences
$$\eqalign{ & \frac{{{V_2}}}{{{V_1}}} = \frac{{{i_2} \times {R_2}}}{{{i_1} \times {R_1}}} = \frac{{{i_2} \times {R_2}}}{{2{i_2} \times 2{R_2}}} = \frac{1}{4}\,\,\left[ {R \propto r} \right] \cr & \therefore \frac{{{V_1}}}{{{V_2}}} = \frac{4}{1} \cr} $$
NOTE
If wires are made of same material, then resistance of coil is proportional to the radius of coil.
i.e., $$R \propto I\,\,{\text{so,}}\,\,R \propto 2\pi r$$

Current in the element $$ = J\left( {2\pi r.dr} \right)$$
Current enclosed by ampere loop of radius $$\frac{a}{2}$$

$$\eqalign{ & I = \int\limits_0^{\frac{a}{2}} {\frac{{{J_0}r}}{a}.} 2\pi r.dr \cr & = \frac{{2\pi {J_0}}}{{3a}}{\left( {\frac{a}{2}} \right)^3} = \frac{{\pi {J_0}{a^3}}}{{12}} \cr} $$
Applying ampere’s law
$$\eqalign{ & B.2\pi .\frac{a}{2} = {\mu _0}.\frac{{\pi {J_0}{a^2}}}{{12}} \cr & \Rightarrow B = \frac{{{\mu _0}{J_0}a}}{{12}} \cr} $$
On putting values, $$B = 10\mu T$$

Radius of circular orbit $$= r$$
No. of rotations per second $$= n$$
i.e., $$T = \frac{1}{n}$$
Magnetic field at its centre, $${B_c} = ?$$
As we know, current
$$i = \frac{e}{T} = \frac{e}{{\left( {\frac{1}{n}} \right)}} = en = {\text{equivalent current}}$$
Magnetic field at the centre of circular orbit,
$${B_c} = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0}ne}}{{2r}}.$$

According to the question, the magnetic field is perpendicular to the direction of motion of charged particle.
Force on the charged particle is given by
$$F = qvB\sin \theta = q\left( {v \times B} \right)$$
where, $$q =$$  charge on charged particle
$$v =$$  velocity of charged particle
$$B =$$  magnetic field
$$\theta =$$  angle between $$v$$ and $$B = {90^ \circ }$$
According to Fleming’s left hand rule, the force acts perpendicular to the velocity of the particle. This force in magnitude remains same but the direction of charged particle goes on changing and always perpendicular to the velocity of the particle, so the particle will move in a circular orbit. The magnetic force does not make any change in its kinetic energy which implies that speed is constant or unchanged.

When a charged particle $$q$$ is moving in a uniform magnetic field $$B$$ with velocity $$v$$ such that angle between $$v$$ and $$B$$ is $$\theta ,$$ then the charge $$q$$ experiences a force which is given by
$$F = q\left( {v \times B} \right) = qvB\sin \theta $$
If $$\theta = {0^ \circ }$$  or $${180^ \circ },$$ then $$\sin \theta = 0$$
$$\therefore F = qvB\sin \theta = 0$$
Since, force on charged particle is non-zero, so angle between $$v$$ and $$B$$ can have any value other than $${0^ \circ }$$ and $${180^ \circ }.$$
NOTE
Force experienced by the charged particle is the Lorentz force.

There is no current inside the pipe. Therefore
$$\eqalign{ & \oint {\overrightarrow B .\overrightarrow {d\ell } = {\mu _0}I} \cr & I = 0 \cr & \therefore B = 0 \cr} $$

According to Maxwell's right handed screw rule, the magnetic field at right hand of wire 1 is perpendicular to the current carrying wire in plane of paper going inwards shown by $$ \otimes .$$ Similarly, the magnetic field at left hand of wire 2 is perpendicular to current carrying wire in plane of paper opposite to first wire shown by $$ \odot .$$ Thus, the two fields are opposite to each other.
Therefore, net magnetic field
$$B = {B_1} - {B_2} = \frac{{{\mu _0}{i_1}}}{{2\pi {r_1}}} - \frac{{{\mu _0}{i_2}}}{{2\pi {r_2}}}$$
At mid-point, $${r_1} = {r_2} = \frac{r}{2} = \frac{5}{2} = 2.5\,cm$$
Hence, $$B = \frac{{{\mu _0}}}{{2\pi }}\left( {\frac{{{i_1}}}{{\frac{r}{2}}} - \frac{{{i_2}}}{{\frac{r}{2}}}} \right)$$
$$\eqalign{ & = \frac{{{\mu _0}}}{{2\pi }}\left( {\frac{5}{{2.5}} - \frac{{2.5}}{{2.5}}} \right) \cr & = \frac{{{\mu _0}}}{{2\pi }}\left( {2 - 1} \right) \cr & = \frac{{{\mu _0}}}{{2\pi }}T \cr} $$

Magnetic Lorentz force, $$F = q\left( {v \times B} \right)$$
$$\eqalign{ & = - 2 \times {10^{ - 6}}\left\{ {\left( {2\hat i + 3\hat j} \right) \times {{10}^6} \times 2\hat j} \right\} \cr & = - 2 \times {10^{ - 6}}\left[ {2 \times 2 \times {{10}^6}\hat k} \right] \cr} $$
$$=8\,N$$  along negative $$Z$$-axis

Clearly, the magnetic fields at a point $$P,$$ equidistant from $$AOB$$  and $$COD$$  will have directions perpendicular to each other, as they are placed normal to each other.
$$\therefore $$ Resultant field, $$B = \sqrt {B_1^2 + B_2^2} $$
$$\eqalign{ & {\text{But}}\,{B_1} = \frac{{{\mu _0}{I_1}}}{{2\pi d}}\,{\text{and}}\,{B_2} = \frac{{{\mu _0}{I_2}}}{{2\pi d}} \cr & \therefore B = \sqrt {{{\left( {\frac{{{\mu _0}}}{{2\pi d}}} \right)}^2}\left( {I_1^2 + I_2^2} \right)} \cr & {\text{or,}}\,B = \frac{{{\mu _0}}}{{2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{\frac{1}{2}}} \cr} $$

$$\eqalign{ & {\text{In}}\,\Delta OAC\,\,\,\,\,\cos {60^ \circ } = \frac{{OC}}{{OA}} \cr & \therefore OC = 2a \times \frac{1}{2} = a \cr} $$
The magnetic field at $$'O'$$ due to
$$\eqalign{ & AB = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{a}\left[ {\sin {{60}^ \circ } - \sin {{30}^ \circ }} \right] \cr & = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{a}\left[ {\frac{{\sqrt 3 }}{2} - \frac{1}{2}} \right] = \frac{{{\mu _0}I}}{{4\pi a}} \times \frac{1}{2}\left( {\sqrt 3 - 1} \right) \cr} $$
The total magnetic field due to all the straight segments of the star is
$$ = \left[ {\frac{{{\mu _0}}}{{4\pi }}\frac{I}{a} \times \frac{1}{2}\left( {\sqrt 3 - 1} \right)} \right] \times 12 = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{a} \times 6\left( {\sqrt 3 - 1} \right)$$