Q: An electron (mass 9 10^ -31 ~kg and charge 1.6 10^ -19 C ) moving with speed c / 100( c = speed of light) is injected into a magnetic field B of magnitude 9 10^ -4 ~T perpendicular to its direction of

For particle To Remain undeflected aligned & F _e=- F _ m \\ & | E |= VB \\ & E = c 100 9 10^ -4 90 \\ & E =27 10^2 ~V / m and \\ & E = B V aligned

Q: A 2 amp current is flowing through two different small circular copper coils having radii ratio 1: 2. The ratio of their respective magnetic moments will be

aligned & Magnetic moment M = I ( ~A )\\ & aligned & M = I ( r ^2 ) \\ & M _1 M _2 = I r_1^2 I r_2^2 = ( r_1 r_2 )^2= ( 1 2 )^2 \\ & M _1 M _2 = 1 4 aligned aligned

Q: A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h / e) where n is an integer, h is Planck's constant and e

Magnetic moment aligned & M=I A=I ( r^2 ) \\ & M= ( e u 2 r ) ( r^2 ) aligned Given B ( r^2 )=n ( h e ) r ^2= h ~B e (2) ( n =1) And when charge moving in external magnetic field then r= m u q B v r = e ~B ~m (3) ( q =e) Put value from equation (2) and (3) in equation (1) aligned & M= ( e v 2 r ) ( r^2 ) \\ & M= e 2 ( e B m ) ( h B e ) \\ & M= e h 2 m aligned

Q: An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60^ with each other. The magnetic moment of this new magnet is :

aligned & I=2 I 2 30^ \\ & = 1 2 \\ & M^ =m / / 2 \\ & =M / 2 aligned

Question 1

The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot whi

Accepted Answer

For a long straight solid wire of radius a carrying a steady current I uniformly distributed across its cross-section, we can find the magnetic field B at a distance r from the axis using Ampere's circuital law: B d l = _0 I_ enclosed . Case 1: Inside the wire (r The current enclosed by an Amperian loop of radius r is I_ enclosed = I ( r^2 a^2 ) = I r^2 a^2 . Applying Ampere's law: B(2 r) = _0 (I r^2 a^2 ) B = ( _0 I 2 a^2 ) r Thus, B r. The graph is a straight line passing through the origin. C

Question 2

A current I_0 flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of magnetic field at the center O of the loop is :

Accepted Answer

From the given figure, the dashed line passing through A, O, and C is straight, indicating that the segments ABC and ADC are semicircles. Let the resistance of segment ADC be R_ ADC = 2R. Given that the resistance of segment ABC is half that of ADC, we have R_ ABC = R. The total current I_0 splits at point A into two parallel branches. Let I_1 be the current through ABC and I_2 be the current through ADC. Since the branches are in parallel, the potential difference across them is equal: I_1 R_ A

Question 3

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 10^ -3 T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil a

Accepted Answer

The magnetic field at the centre of a circular coil is given by: B = _0 N I 2R Substituting the given values (B = 3.14 10^ -3 T, N = 100, R = 0.05 m, _0 = 4 10^ -7 T m/A): 3.14 10^ -3 = 4 10^ -7 100 I 2 0.05 Taking = 3.14: 3.14 10^ -3 = 4 3.14 10^ -5 I 0.1 10^ -3 = 4 10^ -4 I I = 10^ -3 4 10^ -4 = 2.5 A The magnetic moment of the coil is given by: M = N I A = N I ( R^2) M = 100 2.5 3.14 (0.05)^2 M = 250 3.14 0.0025 M = 1.9625 A m ^2 2 A m ^2 Answer: 2.5 A, 2 A m^2

Question 4

Two infinitely long parallel conducting wires A and B carry currents I and 2I, respectively, in the same direction. The wire A has uniform mass per unit length and lies on an insulated floor. The wire

Accepted Answer

The magnetic force per unit length between two parallel wires carrying currents I_1 and I_2 separated by a distance h is given by: f = _0 I_1 I_2 2 h Given I_1 = I and I_2 = 2I, the upward attractive force per unit length on wire A due to wire B is: f = _0 (I)(2I) 2 h = _0 I^2 h The downward gravitational force per unit length on wire A is its weight per unit length: W = g For wire A not to rise from the floor, the upward magnetic force must be less than or equal to the downward gravitational fo

Question 5

An electron (mass 9 10^ -31 ~kg and charge 1.6 10^ -19 C ) moving with speed c / 100( c = speed of light) is injected into a magnetic field B of magnitude 9 10^ -4 ~T perpendicular to its direction of

Accepted Answer

For particle To Remain undeflected aligned & F _e=- F _ m \ & | E |= VB \ & E = c 100 9 10^ -4 90 \ & E =27 10^2 ~V / m and \ & E = B V aligned

Question 6

A 2 amp current is flowing through two different small circular copper coils having radii ratio 1: 2. The ratio of their respective magnetic moments will be

Accepted Answer

aligned & Magnetic moment M = I ( ~A )\ & aligned & M = I ( r ^2 ) \ & M _1 M _2 = I r_1^2 I r_2^2 = ( r_1 r_2 )^2= ( 1 2 )^2 \ & M _1 M _2 = 1 4 aligned aligned

Question 7

A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h / e) where n is an integer, h is Planck's constant and e

Accepted Answer

Magnetic moment aligned & M=I A=I ( r^2 ) \ & M= ( e u 2 r ) ( r^2 ) aligned Given B ( r^2 )=n ( h e ) r ^2= h ~B e (2) ( n =1) And when charge moving in external magnetic field then r= m u q B v r = e ~B ~m (3) ( q =e) Put value from equation (2) and (3) in equation (1) aligned & M= ( e v 2 r ) ( r^2 ) \ & M= e 2 ( e B m ) ( h B e ) \ & M= e h 2 m aligned

Question 8

In a uniform magnetic field of 0.049 T , a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 10^ -6 ~kg ~m ^2. If the magnitude of mag

Accepted Answer

Time period of Oscillation, T=2 I M B aligned & 1 4 =2 9.8 10^ -6 M 0.049 \ & 1 16 =4 ^2 9.8 10^ -6 M 49 10^ -3 \ & M= 4 ^2 9.8 10^ -6 49 10^ -3 16 \ & = 4 ^2 9.8 16 10^ -3 49 \ & =12.8 ^2 10^ -3 10^ -2 10^2 \ & =1280 ^2 10^ -5 Am ^2 \ & aligned

Question 9

An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60^ with each other. The magnetic moment of this new magnet is :

Accepted Answer

aligned & I=2 I 2 30^ \ & = 1 2 \ & M^ =m / / 2 \ & =M / 2 aligned

Question 10

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A . The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4 10^ -7 SI units):

Accepted Answer

The magnitude of magnetic field due to circular coil of N turns is given by aligned & B_C= _0 i N 2 R \ & = 4 10^ -7 7 100 2 0.1 \ & =4.4 10^ -3 ~T \ & =4.4 mT aligned

Question 11

The magnetic potential energy, when a magnetic bar of magnetic moment m is placed perpendicular to the magnetic field B is

Accepted Answer

Potential energy stored in external magnetic field is given by U=- m B Angle between m and B is 90^ U=-m B 90^ =0

Magnetic Effects of Current — Previous Year Questions & Solutions