Magnetic Effects of Current and Magnetism MCQs for NEET — Physics Questions with Answers

Diamagnetic materials do not have a permanent magnetic dipole moment; they are characterized by their ability to create an induced magnetic field in a direction opposite to an externally applied magnetic field. Therefore, the assertion is true (diamagnetic materials can exhibit magnetism in the presence of an external magnetic field), but the reason is false.

For paramagnetic materials, the magnetization increases as the temperature decreases. This is because the thermal agitation of the atoms reduces, allowing the magnetic dipoles to align more easily with the external magnetic field. Hence, the assertion is true, but the reason is false since magnetization does depend on temperature.

When two short magnets are placed on a cork such that their axes are perpendicular, the magnetic forces balance each other out, resulting in no net force or torque acting on the cork. This means the cork will have neither translational nor rotational motion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Both the assertion and reason are true. A current-carrying solenoid produces a magnetic field that is uniform and independent of its length and cross-sectional area. However, the reason does not explain why the magnetic field is independent of these factors.

Metallic bonding is formed due to attraction of valence (free) electrons with the positive ion cores.Their conductivity decreases with rise of temperature.When visible light falls on a metallic crystal, the electrons of atom absorb visible light, so they are opaque to visible light.However, some orbital state.They then return to their normal state, remitting light of same frequency.

Magnetic needle deflect in the presence of a magnetic field produced by the wire.

(b) $M=i\left({\mathrm{\pi r}}^2\right)=\frac{ev}{2\mathrm{\pi r}}\times {\mathrm{\pi r}}^2\Rightarrow \mathrm{M}=\frac{1}{2}\mathrm{evr}$

(a)  Field at the centre  ${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{2\mathrm{\pi in}}{\mathrm{r}}=\frac{{\mathrm{\mu }}_0}{2}·\frac{\mathrm{ni}}{\mathrm{r}}$

      Field at a distance h from the centre

      $$\begin{gathered} {\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}·\frac{2{\mathrm{\pi nir}}^2}{({\mathrm{r}}^2+{\mathrm{h}}^2{)}^{3/2}}=\frac{{\mathrm{\mu }}_0}{2}·\frac{{\mathrm{nir}}^2}{{\mathrm{r}}^3{\left(1+{\displaystyle \frac{{\mathrm{h}}^2}{{\mathrm{r}}^2}}\right)}^{3/2}} \\ ={\mathrm{B}}_1{\left(1+\frac{{\mathrm{h}}^2}{{\mathrm{r}}^2}\right)}^{-3/2} ={\mathrm{B}}_1\left(1-\frac{3}{2}·\frac{{\mathrm{h}}^2}{{\mathrm{r}}^2}\right) \end{gathered}$$

              (By binomial theorem)

     Hence ${\mathrm{B}}_2$ is less than ${\mathrm{B}}_1$ by a fraction $=\frac{3}{2}\frac{{\mathrm{h}}^2}{{\mathrm{r}}^2}$ 

(c)   $i=6.6\times {10}^{15}\times 1.6\times {10}^{-19}=10.5\times {10}^{-4}amp$

        $A={\mathrm{\pi R}}^2=3.142\times {\left(0.528\right)}^2\times {10}^{-2}{\mathrm{m}}^2$

        $\Rightarrow M=iA=10.5\times {10}^{-4}\times 3.142\times {\left(0.528\right)}^2\times {10}^{-20}$

        $=10\times {10}^{-24}units=1\times {10}^{-23}units$

(c) The effective current $i=\frac{q\omega }{2\mathrm{\pi }} and A={\mathrm{\pi r}}^2$ .
      Magnetic moment $M=iA=\frac{1}{2}q\omega r^2$
      Angular moment $L=I\omega =m{r}^2\omega \Rightarrow \frac{M}{L}=\frac{q}{2m}$