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

KEY CONCEPT : When a charged particle is moving at right angles to the magnetic field then a force acts on it which behaves as a centripetal force and moves the particle in circular motion.
$$\eqalign{ & \therefore \frac{{{m_A}v_A^2}}{{{r_A}}} = q.{v_A}B\,\,\,\,\,\therefore \frac{{{m_A}{v_A}}}{{{r_A}}} = qB \cr & \frac{{{m_B}{v_B}}}{{{r_B}}} = qB \cr & \Rightarrow \frac{{{m_A}{v_A}}}{{{r_A}}} = \frac{{{m_B}{v_B}}}{{{r_B}}} \cr & {\text{Since}}\,{r_A} > {r_B} \Rightarrow {m_A}{v_A} > {m_B}{v_B} \cr} $$

Magnetic field at the centre of circular coil carrying current $$i$$ with $$N$$ number of tums is given by
$$B = \frac{{{\mu _0}Ni}}{{2r}}$$
I
$$\eqalign{ & N = 1,L = 2\pi r \cr & \Rightarrow r = \frac{L}{{2\pi }} \cr & \therefore B = \frac{{{\mu _0} \times 1 \times i}}{{2r}} = \frac{{{\mu _0}i}}{{2r}} \cr} $$
II
$$\eqalign{ & N = 2,L = 2 \times 2\pi r' \cr & \Rightarrow r' = \frac{L}{{4\pi }} = \frac{r}{2} \cr & \therefore B' = \frac{{{\mu _0} \times 2 \times i}}{{2r'}} \cr} $$
Putting the value of $${r'}$$
$$B' = \frac{{{\mu _0} \times 2i}}{{2 \times \left( {\frac{r}{2}} \right)}} = \frac{{4{\mu _0}i}}{{2r}} = 4B$$
NOTE
Magnetic field at the centre of circular coil is maximum and decreases as we move away from the centre (on the axis of coil).

When a charged particle moves through a perpendicular magnetic field, then a magnetic force acts on it which changes the direction of particle but does not alter the magnitude of its velocity (i.e., speed).
So, speed of charged particle remains unchanged i.e., of velocity magnitude remains constant.
NOTE
If a charged particle moves at $${45^ \circ }$$ to magnetic field, then path of the particle will be a helix whose circular part has radius according to relation, $$r = \frac{{mv\sin \theta }}{{qB}}.$$

Torque on the solenoid is given by $$\tau = MB\sin \theta $$
where $$\theta $$ is the angle between the magnetic field and the axis of solenoid.
$$\eqalign{ & M = niA \cr & \therefore \tau = niA\,B\sin {30^ \circ } \cr & = 2000 \times 2 \times 1.5 \times {10^{ - 4}} \times 5 \times {10^{ - 2}} \times \frac{1}{2} \cr & = 1.5 \times {10^{ - 2}}N - m \cr} $$

Consider two amperian loops of radius $$\frac{a}{2}$$ and $$2a$$ as shown in the diagram.
Applying Ampere’s circuital law for these loops we get,

$$\oint {B.dL} = {\mu _0}{I_{{\text{enclosed}}}}$$
For the smaller loop
$$\eqalign{ & \Rightarrow B \times 2\pi \frac{a}{2} = {\mu _0} \times \frac{I}{{\pi {a^2}}} \times \pi {\left( {\frac{a}{2}} \right)^2} \cr & = {\mu _0}I \times \frac{1}{4} = \frac{{{\mu _0}I}}{4} \cr} $$
$$ \Rightarrow {B_f} = \frac{{{\mu _0}I}}{{4\pi a}},$$    at distance $$\frac{a}{2}$$ from the axis of the wire.
Similarly, for bigger amperian loop.
$$\eqalign{ & B' \times 2\pi \left( {2a} \right) = {\mu _0}I\,\,\left[ {{\text{total current enclosed by Amperian loop is 2}}} \right] \cr & \Rightarrow B' = \frac{{{\mu _0}I}}{{4\pi a}}, \cr} $$
at distance $$2a$$ from the axis of the wire.
So, ratio of, $$\frac{B}{{B'}} = \frac{{{\mu _0}I}}{{4\pi a}} \times \frac{{4\pi a}}{{{\mu _0}I}} = 1$$

Magnetic field $$\left( B \right)$$ will not apply any force. Only electric field $$E$$ will apply a force opposite to velocity of the electron hence, speed decreases.

If charged particle is moving perpendicular to the direction of $$B,$$ it experiences a maximum force which acts perpendicular to the direction of $$B$$ as well as $$v.$$ Hence, this force will provide the required centripetal force and the charged particle will describe a circular path in the magnetic field of radius $$r$$ and is given by
$$\frac{{m{v^2}}}{r} = qvB$$
Now, $$KE$$  of electron $$ = 10\,eV$$
$$\eqalign{ & \Rightarrow \frac{1}{2}m{v^2} = 10\,eV \cr & \therefore \frac{1}{2} \times \left( {9.1 \times {{10}^{ - 31}}} \right){v^2} = 10 \times 1.6 \times {10^{ - 19}} \cr & \Rightarrow {v^2} = \frac{{2 \times 10 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}} \cr & \Rightarrow {v^2} = 3.52 \times {10^{12}} \cr & {\text{or}}\,\,v = 1.88 \times {10^6}m \cr} $$
Now, radius of circular path,
$$\eqalign{ & r = \frac{{mv}}{{qB}} = \frac{{9.1 \times {{10}^{ - 31}} \times 1.88 \times {{10}^6}}}{{1.6 \times {{10}^{ - 19}} \times {{10}^{ - 4}}}} \cr & = 11\,cm \cr} $$

As the net force on closed loop is equal to zero. So, force on $$QP$$  will be equal and opposite to sum of forces on other 3 sides.

So, from vector laws, $${F_{QP}} = \sqrt {{{\left( {{F_3} - {F_1}} \right)}^2} + F_2^2} $$

Magnetic field on axis of circular coil is
$$\eqalign{ & \frac{{{\mu _0}i{R^2}}}{{2{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}} = \frac{{{\mu _0}i{R^2}}}{{2{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}} = \frac{{{\mu _0}{i_2}{R^2}}}{{2{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}} \cr & x = \sqrt 7 R \cr} $$

Case 1: Magnetic field at $$M$$ due to $$PQ$$ and $$QR$$  is
$${H_1} = \frac{1}{2}\left[ {\frac{{{\mu _0}I}}{{2\pi R}}} \right] + 0 = \frac{{{\mu _0}I}}{{4\pi R}}$$

Case 2: When wire $$QS$$ is joined.
$${H_2}$$ = (Magnetic field at $$M$$ due to $$PQ$$ ) + (magnetic field $$M$$ due to $$QR$$ ) + (Magnetic field at $$M$$ due to $$QS$$ )
$$ = \frac{1}{2}\left[ {\frac{{{\mu _0}I}}{{2\pi R}}} \right] + 0 + \frac{1}{2}\left[ {\frac{{{\mu _0}\frac{I}{2}}}{{2\pi R}}} \right] = \frac{{3{\mu _0}I}}{{8\pi R}}\,\,\,\,\therefore \frac{{{H_1}}}{{{H_2}}} = \frac{2}{3}$$
NOTE : The magnetic field due to an infinitely long wire carrying current at a distance $$R$$ from the end point is half that at a distance $$R$$ from the middle point.