NEET Magnetic Field Due to Circular Current Carrying Wire MCQs & PYQs

Practice NEET Magnetic Field Due to Circular Current Carrying Wire Questions

Question 1:

moderate

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the center of this coil of n turns will be:

(2016 – II)

For a single turn loop, radius $R = \frac{L}{2\pi}$ and $B = \frac{\mu_0 i}{2R}$. When bent into $n$ turns, the new radius is $R' = \frac{R}{n}$. The magnetic field becomes $B' = \frac{\mu_0 n i}{2R'} = n^2 B$.

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Question 2:

moderate

An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has magnitude:

(2015)

The current produced by the revolving electron is $i = qf = ne$. The magnetic field at the center of a circular loop of radius $r$ carrying current $i$ is $B = \frac{\mu_0 i}{2r} = \frac{\mu_0 n e}{2r}$.

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Question 3:

moderate

Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant-magnetic field induction at the center will be:

(2012 Pre)

The magnetic fields due to the two perpendicular coils are $B_1 = \frac{\mu_0 I}{2R}$ and $B_2 = \frac{\mu_0 (2I)}{2R} = \frac{\mu_0 I}{R}$. Since their planes are at right angles, the fields are perpendicular, so $B_{\text{res}} = \sqrt{B_1^2 + B_2^2} = \frac{\sqrt{5}\mu_0 I}{2R}$.

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Question 4:

moderate

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is:

(2011 Mains)

The equivalent current is $i = qf$. The magnetic induction at the center of a circular ring of radius $R$ carrying current $i$ is $B = \frac{\mu_0 i}{2R} = \frac{\mu_0 q f}{2R}$.

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Question 5:

moderate

The magnetic field of given length of wire for single turn coil at its centre is ‘B’ then its value for two turns coil for the same wire is:

(2002)

For a wire of length $L$, radius for single turn is $R = L/(2\pi)$ and for two turns is $R' = L/(4\pi) = R/2$. Magnetic field $B = \frac{\mu_0 N I}{2 R}$. For 2 turns, $B' = \frac{\mu_0 (2) I}{2 (R/2)} = 4 \left(\frac{\mu_0 I}{2R}\right) = 4B$.

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