Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field B with the distance d from the centre of the conductor is correctly represented by the figure:

Graph 1

Graph 2

Graph 3

Graph 4

Step-by-Step Solution

Inside the Conductor ( $d < R$ ): Using Ampere's Circuital Law for a solid cylindrical conductor carrying a uniform current $I$ , the magnetic field $B$ at a distance $d$ inside is derived by considering the current enclosed by a loop of radius $d$ . The enclosed current is $I_{encl} = I(d^2/R^2)$ . Applying $\oint B \cdot dl = \mu_0 I_{encl}$ yields $B(2\pi d) = \mu_0 I (d^2/R^2)$ , which simplifies to $B = \left(\frac{\mu_0 I}{2\pi R^2}\right)d$ . Thus, $B \propto d$ (Linear increase) .
Outside the Conductor ( $d > R$ ): For points outside, the enclosed current is the total current $I$ . Ampere's Law gives $B(2\pi d) = \mu_0 I$ , which simplifies to $B = \frac{\mu_0 I}{2\pi d}$ . Thus, $B \propto \frac{1}{d}$ (Hyperbolic decrease) .
Graph Characteristics: The correct plot must show a straight line passing through the origin for $d < R$ (linear rise) and a curve decreasing towards zero for $d > R$ (rectangular hyperbola). This corresponds to Figure 4.14 in the NCERT text .

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