Moving Charges and Magnetism: NEET previous year questions

From $B = \frac{\mu_0 N i}{2R}$, the current is $i = \frac{2RB}{\mu_0 N} = \frac{2\times5\times10^{-2}\times3.14\times10^{-3}}{4\pi\times10^{-7}\times100} = 2.5$ A. The magnetic moment $M = NiA = Ni\pi R^2 = 100\times2.5\times\pi\times(0.05)^2 \approx 2$ A m$^2$. Only option (4) gives both correct values.

The shunt is in parallel with the galvanometer, so $i_s r_s = i_G R_G$, where $i_s \approx 10$ A and $i_G = 1$ mA. Thus $10\times r_s = 10^{-3}\times100$, giving $r_s = 0.01$ $\Omega$. The other values come from misplacing the factor of ten in the current ratio.

Why the other options are wrong:

- (2) Off by a factor of 10.
- (3) Off by a factor of 100.
- (4) Off by a factor of 10 (too small).

By Ampere's law, inside the wire ($r<a$) only the enclosed fraction of current contributes, giving $B = \dfrac{\mu_0 I r}{2\pi a^2}$, so $B \propto r$ (a straight line rising to a maximum at $r=a$). Outside ($r>a$) the full current is enclosed, giving $B = \dfrac{\mu_0 I}{2\pi r}$, so $B \propto 1/r$. Only graph (1) shows this linear rise followed by a $1/r$ decay.

The current entering at $A$ divides between the two arcs in INVERSE proportion to their resistances, so with $R_{ABC} = \frac{1}{2}R_{ADC}$ the currents are $\frac{2}{3}I_0$ and $\frac{1}{3}I_0$. Each arc is a semicircle ($A$ at the top, $C$ at the bottom), contributing $\frac{\mu_0 I}{4r}$ but in OPPOSITE directions at $O$. The net field is $\frac{\mu_0}{4r}\left(\frac{2}{3}I_0 - \frac{1}{3}I_0\right) = \frac{\mu_0 I_0}{12r}$. NCERT Class 12, Moving Charges and Magnetism (Biot-Savart law).

Why the other options are wrong:

- (2) sends the whole current $I_0$ round one semicircle and ignores the opposing contribution from the other arc
- (3) uses the full-loop result $\frac{\mu_0 I}{2r}$ without splitting the current or accounting for cancellation
- (4) applies the formula for a long straight wire, $\frac{\mu_0 I}{2\pi r}$, instead of an arc

Parallel currents in the SAME direction attract, so wire $B$ pulls wire $A$ upwards with force per unit length $\frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$. Wire $A$ stays on the floor as long as this does not exceed its weight per unit length $\lambda g$, so $\frac{\mu_0 I^2}{\pi h} \le \lambda g$ and $h_{min} = \frac{\mu_0 I^2}{\pi\lambda g}$. NCERT Class 12, Moving Charges and Magnetism (force between two parallel currents).

The magnetic field B inside a solenoid is given by $B = \mu_0 \frac{N}{L} I$. Substituting the values, $B = 4\pi \times 10^{-7} \times \frac{100}{0.5} \times 2.5 = 6.28 \times 10^{-4}\ T$. Chapter: Moving Charges and Magnetism.