Electromagnetic Induction MCQ Questions & Answers in Electrostatics and Magnetism | Physics

EMF induced in an element of length $$dx$$ at a distance $$x$$ from wire $$= Bvdx$$
∴ Total EMF induced in the rod
$$\eqalign{ & E = \int\limits_{0.01}^{0.2} {Bv} \,dx = \int\limits_{0.01}^{0.2} {\frac{{{\mu _0}iv}}{{2\pi x}}} dx = \frac{{{\mu _0}iv}}{{2\pi }}\int\limits_{0.01}^{0.2} {\frac{1}{x}} dx \cr & E = \frac{{{\mu _0}iv}}{{2\pi }}\left[ {{{\log }_e}x} \right]_{0.01}^{0.2} = \frac{{{\mu _0}iv}}{{2\pi }}\left[ {{{\log }_{10}}\left( {0.2} \right) - {{\log }_{10}}\left( {0.01} \right)} \right] \times 2.303 \cr & E = \frac{{4\pi \times {{10}^{ - 7}} \times 5 \times 10}}{{2\pi }}\left[ {1.301} \right] \times 2.303 \cr & = 2.99 \times {10^{ - 5}}V \cr & \approx 30\,\mu V \cr} $$

When the magnetic flux linked with a coil changes, an induced emf acts in the coil which is given by $$e = - \frac{{d\phi }}{{dt}}$$
The magnetic flux linked with a coil carrying a current $$i,$$ is proportional to $$i.$$
$$\eqalign{ & {\text{or}}\,\,\phi \propto i\,\,{\text{or}}\,\,\phi = Li \cr & \therefore e = - \frac{{d\phi }}{{dt}} = - L\frac{{di}}{{dt}} \cr} $$
The work done in maintaining the current for time $$dt$$
$$ = - ei\,dt = L\frac{{di}}{{dt}}i\,dt$$
and the total work done while the current $${i_0}$$ is being established
$$\eqalign{ & W = \int_0^t {L\frac{{di}}{{dt}}} i\,dt = \int_0^{{i_0}} {Li} \,dt \cr & = \frac{1}{2}Li_0^2 \cr} $$
Thus, an inductor may store energy in its magnetic field.
NOTE
The expression $$\left( {W = \frac{1}{2}Li_0^2} \right)$$   reminds us of $$\frac{1}{2}m{v^2}$$  for mechanical kinetic energy of a particle of mass $$m,$$ and shows that $$L$$ is analogus to $$m$$ (i.e. $$L$$ is electrical inertia, which opposes the growth and decay of current in circuit).

Let radius of the loop is $$r$$ at any time $$t$$ and in further time $$dt,$$  radius increases by $$dr.$$
The change in flux : $$d\phi = \left( {2\pi rdr} \right)B$$

$$\eqalign{ & \Rightarrow e = \frac{{d\phi }}{{dt}} = 2\pi r\left( {\frac{{dr}}{{dt}}} \right)\frac{k}{r} \cr & \Rightarrow e = 2\pi ck\left( {{\text{constant}}} \right)\,\,\left[ {\because \frac{{dr}}{{dt}} = c,B = \frac{k}{r}} \right] \cr} $$
The change in flux : $$d\phi = \left( {2\pi rdr} \right)B$$

$$\eqalign{ & \phi = BA\cos \theta = 2.0 \times 0.5 \times \cos {60^ \circ } \cr & = \frac{{2.0 \times 0.5}}{2} \cr & = 0.5\,{\text{weber}}{\text{.}} \cr} $$

Given, Number of turns of solenoid, $$N = 1000.$$
Current, $$I = 4A$$
Magnetic flux, $${\phi _B} = 4 \times {10^{ - 3}}Wb$$
$$\because $$ Self inductance of solenoid is given by
$$L = \frac{{{\phi _B}.N}}{I}\,.......\left( {\text{i}} \right)$$
Substitute the given values in Eq. (i), we get
$$L = \frac{{4 \times {{10}^{ - 3}} \times 1000}}{4} = 1\,H$$

A long solenoid is that whose length is very large as compared to its radius of cross-section. If $$N$$ is total number of turns in the solenoid, $$A$$ is area of each turn of the solenoid and $$l$$ is length of solenoid, then self-inductance of solenoid is given by
$$\eqalign{ & L = \frac{{{\mu _0}{N^2}A}}{l} \Rightarrow L = {\mu _0}{n^2}Al\,\,\,\left( {n = {\text{number of turns per unit length}}} \right) \cr & So,\,\,L \propto {n^2} \cr} $$
When $${n^2}$$ is doubled, $$L$$ becomes 4 times.

$$\eqalign{ & \left| e \right| = B\frac{{dA}}{{dt}} \cr & \left| e \right| = B\frac{{\left( {{A_2} - {A_1}} \right)}}{{\left( {{t_2} - {t_1}} \right)}} \cr & \Rightarrow \left| e \right| = \frac{{1\left( {\frac{\pi }{4} - 0} \right)}}{{\left( {1 - 0} \right)}} \cr & e = \frac{\pi }{4}V \cr} $$

KEY CONCEPT :
Time constant of $$R - C\,{\text{circuit}}\,{\text{is}}\,\tau = {R_{eq}}{C_{eq}}$$
$$\eqalign{ & \left( {\text{i}} \right)\,\,{\tau _1} = \left( {2 + 1} \right)\left( {2 + 4} \right) = 18\mu {\text{s}} \cr & \left( {{\text{ii}}} \right)\,{\tau _2} = \left( {\frac{{2 \times 1}}{{2 + 1}}} \right)\left( {\frac{{2 \times 4}}{{2 + 4}}} \right) = \frac{8}{9}\mu {\text{s}} \cr & \left( {{\text{iii}}} \right)\,{\tau _3} = \left( {\frac{{2 \times 1}}{{2 + 1}}} \right) \times \left( {4 + 2} \right) = 4\mu {\text{s}} \cr} $$

Component of weight along the inclined plane $$ = mg\sin \theta $$
Again, $$F = BI\ell = B\frac{{B\ell v}}{R}\ell = \frac{{{B^2}{\ell ^2}v}}{R}$$
Now, $$\frac{{{B^2}{\ell ^2}v}}{R} = mg\sin \theta \,{\text{or}}\,v = \frac{{mgR\sin \theta }}{{{B^2}{l^2}}}$$

According to question, two identical charged spheres suspended from a common point by two massless strings of length $$L.$$

$$\because {\text{In,}}\,\,\Delta ABC\,\tan \theta = \frac{F}{{mg}}\,\,{\text{or}}\,\theta \frac{F}{{mg}} = \tan \theta \,......\left( {\text{i}} \right)$$
Since, the charges begins to leak from both the spheres at a constant rate. As a result, the spheres approach each other with velocity $$v.$$
Therefore, Eq. (i) can be rewritten as
$$\eqalign{ & \frac{{K{q^2}}}{{{x^2}mg}} = \frac{{\frac{x}{2}}}{{\sqrt {{l^2} - \frac{{{x^2}}}{4}} }} \cr & \Rightarrow \frac{{K{q^2}}}{{{x^2}mg}} = \frac{x}{{2l}}\,\,{\text{or}}\,\,{q^2} \propto {x^3} \cr & \Rightarrow q \propto {x^{\frac{3}{2}}} \cr & \Rightarrow \frac{{dq}}{{dt}} \propto \frac{{d\left( {{x^{\frac{3}{2}}}} \right)}}{{dx}} \cdot \frac{{dx}}{{dt}} \cr & \Rightarrow \frac{{dq}}{{dt}} \propto {x^{\frac{1}{2}}} \cdot v \cr & \Rightarrow v \propto \frac{1}{{{x^{\frac{1}{2}}}}}\,\,{\text{or}}\,\,v \propto {x^{ - \frac{1}{2}}} \cr} $$