Class 12th

A magnetic pole will repel or attract magnetic sheet so force is need.

B. If sheet is non-magnetic, no force needed.

C. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed.

D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet.

$f_0=140 cm$ and $f_e=5 cm$

For distant object,

$m=\frac{f_0}{f_e}=\frac{140}{5}=28$

Given  $V^{\prime }=V=$ Constant

 (i) $C^{\prime }=\frac{{\epsilon }_{0}A}{d^{\prime }},C=\frac{{\epsilon }_{0}A}{d}$

$d^{\prime }<d$

$C^{\prime }>C$
Hence, final capacitance greater than initial capacitance,

(ii)  $U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2$

$U=\frac{1}{2}C{V}^2$

$U^{\prime }>U$

Hence final energy is greater than initial energy

(iii)  $\frac{Q^{\prime }}{V^{\prime }}=C^{\prime }$ and $\frac{Q}{V}=C$

$\frac{Q^{\prime }}{V^{\prime }}\ne \frac{Q}{V}$

(iv) Product of charge and voltage

$X^{\prime }=Q^{\prime }V=C^{\prime }{V}^2$

$X=QV=C{V}^2$

$X^{\prime }>X$

$M=ml$

$\Delta I=2\frac{l}{2}sin30^{?}$

$=\frac{l}{2}$

$M^{\prime }=mI/2$

$=M/2$

Power Consumed $=P=\frac{V^2}{R}$

$\frac{P_A}{P_B}=\frac{R_B}{R_A}$

$R_A=2R_B$

For Series Combination

$P_S=\frac{V^2}{3R_B}$

For Parallel Combination

$P_P=\frac{3V^2}{2R_B}$

$\frac{P_S}{P_P}=\frac{2}{9}$

In option (1),

$\frac{10}{15}=\frac{10}{5+R_D}$

The diode can conduct and have resistance  $R_D=10\Omega$  because diode have dynamic resistance. In that case bridge will be balanced.

The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field.

Capacitive Reactance $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2*3.14*50*10*10^{-6}}$

$=\frac{1000}{3.14}$

$V_{rms}=210 V$

$i_{rms}=\frac{V_{rms}}{X_C}=\frac{210}{X_C}$

$Peak current =\sqrt[]{2}{i}_{ms}=\sqrt[]{2}*\frac{210}{1000}*3.14=0.932$

$?0.93 A$

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$

(Material)                      (Susceptibility ( $\chi$  )

Diamagnetic                 (II) $0>\chi \ge -1$

Ferromagnetic  (III) $\chi ?1$

Paramagnetic                (IV) $0<\chi <\epsilon$

Non-magnetic               (I) $\chi =0$