NEET Magnetic Effects of Current MCQs & PYQs

Practice NEET Magnetic Effects of Current Questions

Question 1:

moderate

Given below are two statements

Statement I : Biot-Savart’s law gives us the expression for the magnetic field strength of an infinitesimal current element ($I d \vec{l}$) of the current carrying conductor only.

Statement II : Biot-Savart’s law is analogous to Coulomb’s inverse square law of charge $q$, with the former being related to the field produced by a scalar source, $Id \vec{l}$, while the latter being produced by a vector source, $q$.

In light of above statements choose the most appropriate answer from the options given below

Statement I is correct as Biot-Savart's law defines the magnetic field for a current element $I d \vec{l}$. Statement II is incorrect because $I d \vec{l}$ is a vector source while charge $q$ is a scalar source, reversing the description.

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Question 2:

easy

Tesla is the unit of

(1997, 88)

The SI unit of magnetic field (magnetic induction) is Tesla ($ \text{T} $), defined as one Weber per square meter ($ \text{Wb/m}^2 $).

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Question 3:

easy

The magnetic field at a distance r from a long wire carrying current i is 0.4 tesla. The magnetic field at a distance 2r is

(1992)

Magnetic field due to a long straight wire is inversely proportional to distance ($B \propto 1/r$). When distance is doubled from $r$ to $2r$, the magnetic field is halved. Therefore, the new magnetic field is $0.4 / 2 = 0.2\text{ tesla}$.

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Question 4:

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 5:

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 6:

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 7:

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 8:

moderate

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of potential difference applied across them so that the magnetic field at their center is the same?

(2006)

Magnetic field $B = \frac{\mu_0 N I}{2 R}$. Since $B$ and length of wire are same, $N_1 R_1 = N_2 R_2$. With $R_1 = 2R_2$, $N_1 = N_2 / 2$. The potential difference ratio $\frac{V_1}{V_2} = \frac{I_1 R_{\text{wire}, 1}}{I_2 R_{\text{wire}, 2}} = 4$.

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Question 9:

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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Question 10:

moderate

From Ampere’s circuital law for a long straight wire of circular cross section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:

(2022)

Inside a uniform cylindrical wire, the magnetic field is directly proportional to the radius ($B \propto r$), increasing linearly. Outside the wire, the magnetic field is inversely proportional to the distance ($B \propto 1/r$). Therefore, option D is the correct choice.

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