1. ⇒ (NEET 2023)

A very long conducting wire is bent in a semi-circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by :

NEET 2023 Physics - Moving Charges and Magnetism Question 3 English

A. $\frac{μ_{0} i}{4 R}$ pointed away from the page

B. $\frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ pointed away from page

C. $\frac{μ_{0} i}{4 R} [1 - \frac{2}{π}]$ pointed into the page

D. $\frac{μ_{0} i}{4 R}$ pointed into the page

Correct Answer is Option (B)

$B = \frac{μ_{0}}{4 π} \frac{I}{R} (π) - \frac{μ_{0}}{4 π} \frac{2 I}{R}$

$= \frac{μ_{0} I}{4 R} [1 - \frac{2}{π}]$ outward i.e. away from page.

2. ⇒ (NEET 2022 Phase 2)

The shape of the magnetic field lines due to an infinite long, straight current carrying conductor is

A. a plane

B. a straight line

C. circular

D. elliptical

Correct Answer is Option (C)

NEET 2022 Phase 2 Physics - Moving Charges and Magnetism Question 5 English Explanation

From the right hand curl rule, the shape of magnetic field lines due to long current carrying wire is circular (concentric circles)

3. ⇒ (NEET 2022 Phase 2)

The magnetic field on the axis of a circular loop of radius 100 cm carrying current $I = \sqrt{2} A$ , at point 1 m away from the centre of the loop is given by :

A. $6.28 \times 10^{- 4}$ T

B. $3.14 \times 10^{- 7}$ T

C. $6.28 \times 10^{- 7}$ T

D. $3.14 \times 10^{- 4}$ T

Correct Answer is Option (B)

Magnetic field at the axis of a current carrying circular loop

$| B | = \frac{μ_{0} i a^{2}}{2 {[a^{2} + d^{2}]}^{3 / 2}} \Rightarrow | B | = \frac{4 π \times 10^{- 7} \times \sqrt{2} \times 1}{2 {[1 + 1]}^{3 / 2}}$

$| B | = \frac{4 π \times 10^{- 7} \times \sqrt{2}}{2. {(2)}^{3 / 2}} = \frac{4 π \times 10^{- 7}}{{(2)}^{3 / 2} \times {(2)}^{1 / 2}} = 3.14 \times 10^{- 7}$ T

4. ⇒ (NEET 2022 Phase 1)

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 (Idl) of a 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, Idl 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.

A. Both Statement I and Statement II are correct

B. Both Statement I and Statement II are incorrect

C. Statement I is correct and Statement II is incorrect

D. Statement I is incorrect and Statement II is correct

Correct Answer is Option (C)

According to Biot-Savart's law $d \vec{B} = \frac{μ_{0}}{4 π} \frac{I d \vec{l} \times \vec{r}}{r^{3}}$ which is applicable for infinitesimal element. It is analogous to Coulomb's law, where $I d \vec{l}$ is vector source and electric field is produced by scalar source q. Here statement I is correct and statement II is incorrect.

5. ⇒ ( NEET 2016 Phase 2)

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

A. nB

B. n²B

C. 2nB

D. 2n²B

Correct Answer is Option (B)

For One turn loop :

B = $\frac{μ_{0} i}{2 r}$

For n turn loop :

r = $\frac{R}{n}$

B' = $\frac{μ_{0} n i}{2 \frac{R}{n}}$

= $(\frac{μ_{0} i}{2 R}) n^{2}$

= n²B

6. ⇒ (AIPMT 2015 Cancelled Paper)

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

A. $\frac{μ_{0} n^{2} e}{r}$

B. $\frac{μ_{0} n e}{2 r}$

C. $\frac{μ_{0} n e}{2 π r}$

D. Zero

Correct Answer is Option (B)

Current in the orbit, $I = \frac{e}{T}$

$I = \frac{e}{(2 π / ω)} = \frac{ω e}{2 π}$

$= \frac{(2 π n) e}{2 π} = n e$

Magnetic field at centre of current carrying circular coil is given by

$B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} n e}{2 r}$

7. ⇒ (AIPMT 2015 Cancelled Paper)

A wire carrying current $I$ has the shape shown in adjoining figure.

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 62 English
Linear parts of the wire are very long and parallel to X-axis while semicircular protion of radius R is lying in Y-Z plane. Magtnetic field at pont $O$ is

A. $\vec{B} = - \frac{μ_{0} I}{4 π R} (π \hat{i} + 2 \hat{k})$

B. $\vec{B} = \frac{μ_{0} I}{4 π R} (π \hat{i} - 2 \hat{k})$

C. $\vec{B} = \frac{μ_{0} I}{4 π R} (π \hat{i} + 2 \hat{k})$

D. $\vec{B} = - \frac{μ_{0} I}{4 π R} (π \hat{i} - 2 \hat{k})$

Correct Answer is Option (A)

Magnetic field due to segment ‘1’

$\vec{B_{1}} = \frac{μ_{0} I}{4 π R} [\sin 90^{\circ} + \sin 0^{\circ}] (- \hat{k})$

$= \frac{- μ_{0} I}{4 π R} (\hat{k}) = \vec{B_{3}}$

Magnetic field due to segment 2

$\vec{B_{2}} = \frac{μ_{0} I}{4 R} (- \hat{i}) = \frac{- μ_{0} I}{4 π R} (- π \hat{i})$

AIPMT 2015 Cancelled Paper Physics - Moving Charges and Magnetism Question 62 English Explanation
$∴ \vec{B}$ at centre

$\vec{B_{c}} = \vec{B_{1}} + \vec{B_{2}} + \vec{B_{3}}$

$= \frac{- μ_{0} I}{4 π R} (π \hat{i} + 2 \hat{k})$

8. ⇒ (AIPMT 2014)

Two identical long conducting wires $A O B$ and $C O D$ are placed at right angle to each other, with one above other such that $O$ is their common point for the two. The wires carry $I$ ₁ and $I$ ₂ currents, respectively. Point $P$ is lying at distance f from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $P$ will be

A. $\frac{μ_{0}}{2 π d} (\frac{I_{1}}{I_{2}})$

B. $\frac{μ_{0}}{2 π d} (I_{1} + I_{2})$

C. $\frac{μ_{0}}{2 π d} (I_{1}^{2} - I_{2}^{2})$

D. $\frac{μ_{0}}{2 π d} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

Correct Answer is Option (D)

AIPMT 2014 Physics - Moving Charges and Magnetism Question 59 English Explanation
$B_{1} = \frac{μ_{0} I_{1}}{2 π d} (- \hat{j})$

$B_{2} = \frac{μ_{0} I_{2}}{2 π d} (- \hat{i})$

$B = \sqrt{B_{1}^{2} + B_{2}^{2}}$

$= \frac{μ_{0}}{2 π d} \times \sqrt{I_{1}^{2} + I_{2}^{2}}$

9. ⇒ ( AIPMT 2012 Prelims)

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 2 $I$ , respectively. The resultant magnetic field induction at the centre will be

A. $\frac{\sqrt{5} μ_{0} I}{2 R}$

B. $\frac{\sqrt{5} μ_{0} I}{R}$

C. $\frac{μ_{0} I}{2 R}$

D. $\frac{μ_{0} I}{R}$

Correct Answer is Option (A)

AIPMT 2012 Prelims Physics - Moving Charges and Magnetism Question 55 English Explanation
Magnetic field induction due to vertical loop at the centre O is

$B_{1} = \frac{μ_{0} I}{2 R}$

It acts in horizontal direction.

Magnetic field induction due to horizontal loop at the centre O is

$B_{2} = \frac{μ_{0} 2 I}{2 R}$

It acts in vertically upward direction.

As B₁ and B₂ are perpendicular to each other, therefore the resultant magnetic field induction at the centre O is

$B_{n e t} = \sqrt{B_{1}^{2} + B_{2}^{2}}$

$= \sqrt{{(\frac{μ_{0} I}{2 R})}^{2} + {(\frac{μ_{0} 2 I}{2 R})}^{2}}$

$B_{n e t} = \frac{μ_{0} I}{2 R} \sqrt{{(1)}^{2} + {(2)}^{2}} = \frac{\sqrt{5} μ_{0} I}{2 R}$

10. ⇒ (AIPMT 2011 Mains)

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

A. $\frac{μ_{0} q f}{2 π R}$

B. $\frac{μ_{0} q f}{2 R}$

C. $\frac{μ_{0} q}{2 f R}$

D. $\frac{μ_{0} q}{2 π f R}$

Correct Answer is Option (B)

When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is

I = $\frac{q}{T}$ = qf

Magnetic field at the centre of the ring is

B = $\frac{μ_{0} I}{2 R}$ = $\frac{μ_{0} q f}{2 R}$

⇒Moving Charges And Magnatism

Topic 1 : Biot Sarevts Law 1