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

When magnetic field is perpendicular to motion of charged particle, then particle performs circular motion.
So, centripetal force = magnetic force
i.e. $$\frac{{m{v^2}}}{R} = Bqv\,\,{\text{or}}\,\,R = \frac{{mv}}{{Bq}}$$
Further, time period of the motion
$$T = \frac{{2\pi R}}{v} = \frac{{2\pi \left( {\frac{{mv}}{{Bq}}} \right)}}{v}\,\,{\text{or}}\,\,T = \frac{{2\pi m}}{{Bq}}$$

If $${F_4}$$ is the force on segment $$QP,$$  then
$$\eqalign{ & {{\vec F}_1} + {{\vec F}_2} + {{\vec F}_3} + {{\vec F}_4} = 0 \cr & {\text{and}}\,{{\vec F}_4} = - \left( {{{\vec F}_1} + {{\vec F}_2} + {{\vec F}_3}} \right) \cr & {\text{or}}\,\,{F_4} = \sqrt {{{\left( {{F_3} - {F_1}} \right)}^2} + F_2^2} . \cr} $$

Let us compute the magnetic field due to any one segment :

$$\eqalign{ & B = \frac{{{\mu _0}i}}{{4\pi \left( {d\sin \alpha } \right)}}\left( {\cos {0^ \circ } + \cos \left( {{{180}^ \circ } - \alpha } \right)} \right) \cr & = \frac{{{\mu _0}I}}{{4\pi \left( {d\sin \alpha } \right)}}\left( {1 - \cos \alpha } \right) \cr & = \frac{{{\mu _0}I}}{{4\pi d}}\tan \frac{\alpha }{2} \cr} $$
Resultant field will be
$${B_{{\text{net}}}} = 2B = \frac{{{\mu _0}I}}{{2\pi d}}\tan \frac{\alpha }{2} \Rightarrow K = \frac{{{\mu _0}I}}{{2\pi d}}$$

$${B_{{\text{centre}}}} = \frac{{n \cdot {\mu _0}i}}{{2R}}\,\,\left( {{\text{For a circular coil}}} \right)$$
where, $$n$$ : Number of turns in circular coil

$$\eqalign{ & B = \frac{{{\mu _0}i}}{{2R}} = \frac{{{\mu _0}i\left( {2\pi } \right)}}{{2\left( l \right)}} = \frac{{{\mu _0}\pi i}}{l} \cr & = \frac{{{\mu _0}ni}}{{2\left( {\frac{l}{{2n\pi }}} \right)}} = \frac{{{n^2}{\mu _0}\pi i}}{l} = {n^2}B \cr} $$

Work done in moving the conductor is,
$$\eqalign{ & W = \int_0^2 F dx \cr & = \int_0^2 3 .0 \times {10^{ - 4}}{e^{ - 0.2x}} \times 10 \times 3dx \cr} $$

$$\eqalign{ & = 9 \times {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}} dx \cr & = \frac{{9 \times {{10}^{ - 3}}}}{{0.2}}\left[ { - {e^{ - 0.2 \times 2}} + 1} \right]B = 3.0 \times {10^{ - 4}}{e^{ - 0.2x}}\,\,\left( {{\text{By exponential function}}} \right) \cr & = \frac{{9 \times {{10}^{ - 3}}}}{{0.2}} \times \left[ {1 - {e^{ - 0.4}}} \right] \cr & = 9 \times {10^{ - 3}} \times \left( {0.33} \right) = 2.97 \times {10^{ - 3}}J \cr} $$
Power required to move the conductor is,
$$\eqalign{ & P = \frac{W}{t} \cr & P = \frac{{2.97 \times {{10}^{ - 3}}}}{{\left( {0.2} \right) \times 5 \times {{10}^{ - 3}}}} = 2.97W \cr} $$

When a rectangular coil is suspended in a uniform magnetic field of induction $$B,$$ then forces acting on coil forms a couple, the effect of which is to rotate the coil. Torque is given by
$$\tau = M \times B = \left| M \right|\left| B \right|\sin \theta $$
where, $$M =$$  magnitude of the magnetic dipole moment of the rectangular current loop and $$\theta $$ is angle between $$M$$ and $$B.$$

As from question

The point $$P$$ is lying at a distance $$d$$ along the $$Z$$-axis.
As magnetic field $${B_1}$$ is given by $$ = \frac{{{\mu _0}}}{{2\pi }}\frac{{{I_1}}}{d}$$
and magnetic field $${B_2}$$ is given by $$ = \frac{{{\mu _0}}}{{2\pi }}\frac{{{I_2}}}{d}$$
$${B_1}$$ and $${B_2}$$ are $$ \bot $$ to each other
So, $${B_{{\text{net}}}}$$ is given by
$$\eqalign{ & {B_{{\text{net}}}} = \sqrt {B_1^2 + B_2^2} , \cr & {B_{{\text{net}}}} = \frac{{{\mu _0}}}{{2\pi }}\frac{1}{d}{\left( {I_1^2 + I_2^2} \right)^{\frac{1}{2}}} \cr} $$

$$\eqalign{ & IgG = \left( {I - Ig} \right)s \cr & \therefore {10^{ - 3}} \times 100 = \left( {10 - {{10}^{ - 3}}} \right) \times S \cr & \therefore S \approx 0.01\Omega \cr} $$

At the centre of current carrying, circular coil, the magnetic field is,
$$B = \frac{{{\mu _0}Ni}}{{2r}}$$
where, $$N =$$  number of turns in the coil
$$i =$$  current flowing
$$r =$$  radius of the coil
Given, $$N = 1000,i = 0.1\,A,r = 0.1\,m$$
Substituting the values, we have
$$\eqalign{ & B = \frac{{4\pi \times {{10}^{ - 7}} \times 1000 \times 0.1}}{{2 \times 0.1}} \cr & = 2\pi \times {10^{ - 4}} \cr & = 6.28 \times {10^{ - 4}}T \cr} $$

Magnetic field can never increase the energy of a charged particle so, its kinetic energy will remain same.