Q 1 :

Two insulated circular loops A and B of radius 'a' carrying a current of 'I' in the anti-clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be ______.                [2024]

  • 2μ0I2a

     

  • μ0I2a

     

  • 2μ0Ia

     

  • 2μ0Ia

     

(1)

BA=μ0I2a

BB=μ0I2a   Bnet=2μ0I2a

 



Q 2 :

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R1=2πm and R2=4πm, carrying current I=4A as per figure given below is α×10-7T. The value of α is ______. (Centre O is common for all segments)               [2024]



(3)

μ0i2R2(π2π)+μ0i2R1(π2π)

(μ0i4R2+μ0i4R1)

4π×10-7×44×4π+4π×10-7×44×2π

=3×10-7=α×10-7

α=3



Q 3 :

Two circular coils P and Q of 100 turns each have same radius of π cm. The currents in P and R are 1A and 2A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is xmT, where x= ____ .

[Use μ0=4π×10-7 TmA-1]                     [2024]



(20)

BP=μ0Ni12r=μ0×1×1002π=2×10-3 T

BQ=μ0Ni22r=μ0×2×1002π=4×10-3 T

Bnet=BP2+BQ2=20 mT

x=20



Q 4 :

An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure.

The magnitude of magnetic field at the origin O of the arc is given by:          [2025]

  • μ04πIa[π2+1]

     

  • μ04πIa[3π2+1]

     

  • μ02πIa[π2+2]

     

  • μ04πIa[3π2+2]

     

(2)

B1=μ0I4πa

B2=μ04πIa(3π2)

B3=0

B=μ04πIa(1+3π2)



Q 5 :

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be          [2025]

  • [a4,a2]

     

  • [a2,2a]

     

  • [a2,3a]

     

  • [a4,2a]

     

(2)

Maximum possible magnetic field on the surface

Bmax=μ0I2πa  Bmax2=μ0I4πa

It can be obtained inside as well as outside the wire.

For inside, μ0I4πa=μ0Ir2πa2  r=a2

For outside, μ0I4πa=μ0I2πr  r=2a

Correct answer [a2,2a]



Q 6 :

Let B1 be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let B2 be the magnitude of magnetic field at an axial distance 'x' from the centre. For x : R=3 : 4, B2B1 is:          [2025]

  • 4 : 5

     

  • 16 : 25

     

  • 64 : 125

     

  • 25 : 16

     

(3)

The magnitude of magnetic field at centre of a circular coil, B1=μ0i2R

The magnitude of magnetic field at an axial distance x=3R4 from the centre.

B2=μ0iR22(R2+x2)3/2=μ0iR22(5R4)3=64125(μ0i2R)

 B2B1=64125



Q 7 :

A loop ABCDA, carrying current I = 12A, is placed in a plane, consists of two semi-circular segments of radius R1=6 πm and R2=4 πm. The magnitude of the resultant magnetic field at center O is k×107 T. The value of k is _______ (Given μ0=4π×107 T mA1)          [2025]



(1)

Magnetic field due to AB and CD = 0

B=μ0I2r(θ2π) for an arc

For semi-circule B=μ0I4r

Bnet=μ0I4{4π}μ0I4{6π}

        =μ0I4π{1416}

        =107×12×224=107 T  k=1



Q 8 :

Match the Column – I with Column – II:  [2023]

Choose the correct answer from the options given below:

  • A – I; B – III; C – IV; D – II

     

  • A – III; B – IV; C – I; D – II

     

  • A – II; B – I; C – IV; D – III

     

  • A – III; B – I; C – IV; D – II

     

(4)

AB0=μ04πir+μ0i2r+μ04πir

              =μ02πir+μ0i2r

B0=μ0i2πr(π-1)

BB0=μ04πir+μ04πirπ+μ04πirπ

            =(2μ04πir+μ04πirπ)

B0=μ0i4πr(2+π)

CB0=μ04πir+μ04πirπ+0

B0=μ04πir(1+π)

DB0=μ04πirπ=μ04πir

A-III;  B-I;  C-IV;  D-II



Q 9 :

A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from +ve z direction and lying in xy plane as shown in figure. The plot of j^ component of magnetic field (By) at a distance 'a' (less than radius of the coil) and on yz plane vs z coordinate looks like             [2023]

  •  

  •  

  •  

  •  

(3)

By=0 in the plane of the coil

By is opposite to each other at -z and +z positions.



Q 10 :

The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn. The magnetic induction at the centre of the coil by the same current will be             [2023]

  • 8 T

     

  • 4 T

     

  • 2 T

     

  • 16 T

     

(3)

B=μ0i2R×4  and  B'=μ0i2R'

For R'=4RB'=μ0i8R

B'B=116  B'=2T



Q 11 :

A long conducting wire having a current I flowing through it is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is             [2023]

  • N:n

     

  • n2:N2

     

  • N2:n2

     

  • n:N

     

(3)

l=(2πr)n

r(ln)

B=n(μ0i2r)(μ0i2L)n2

B1B2=(N2n2)



Q 12 :

Find the magnetic field at the point P in figure. The curved portion is a semicircle connected to two long straight wires.             [2023]

  • μ0i2r(1+2π)

     

  • μ0i2r(1+1π)

     

  • μ0i2r(12+12π)

     

  • μ0i2r(12+1π)

     

(3)

BP=(μ0i4r+μ0i4πr)=μ0i2r(12+12π)



Q 13 :

As shown in the figure, a long straight conductor with a semicircular arc of radius π10m is carrying current I=3 A. The magnitude of the magnetic field at the center O of the arc is (The permeability of the vacuum =4π×10-7 NA-2).           [2023]

  • 6μT

     

  • 1μT

     

  • 4μT

     

  • 3μT

     

(4)

BC=μ0I4πR(π)  (B at centre of circular arc)

=μ0I4R=4π×10-7×34×π10=3×10-6 T=3μT



Q 14 :

The ratio of magnetic field at the centre of a current carrying coil of radius r to the magnetic field at distance r from the centre of the coil on its axis is x:1. The value of x is _______.                   [2023]



(8)

Magnetic field at centre (B1)=μ0I2r

Magnetic field on axis =μ0Ir22(r2+d2)3/2

Value of d=r (given)

B2=μ0I42r

B1B2=μ0I2r×42rμ0I=221=81

x=8



Q 15 :

Two identical circular wires of radius 20 cm and carrying current 2 A are placed in perpendicular planes as shown in the figure. The net magnetic field at the centre of the circular wires is ________ ×10-8 T.                      [2023]

(Take π=3.14)



(628)

Magnetic field BC at center =μ0i2r

BC=4π×10-72×0.2×2 T

Net magnetic field is

BC2=4π×10-7×22×0.2×2 T=2π×10-6 T

             =200π×10-8 T=628×10-8 T



Q 16 :

A straight wire carrying a current of 14 A is bent into a semicircular arc of radius 2.2 cm as shown in the figure. The magnetic field produced by the current at the centre (O) of the arc is _______ ×10-4 T.               [2023]



(2)

Bat O=μ0I4R=4π×10-7×144×2.2×10-2=2×10-4 T



Q 17 :

Two identical circular loops P and Q each of radius r are lying in parallel planes such that they have common axis. The current through P and Q are I and 4I respectively in clockwise direction as seen from O. The net magnetic field at O is:                    [2026]

  • μ0I42r towards Q

     

  • 3μ0I42r towards Q

     

  • 3μ0I42r towards P

     

  • μ0I42r towards P

     

(2)

Bnet=B1-B2

=4μ0iR22(R2+R2)3/2-μ0iR22(R2+R2)3/2

=3μ0i42R



Q 18 :

An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :   [2026]

  • -μ02πIr(π-1)i^

     

  • μ02πIr(π-1)i^

     

  • -μ02πIr(π+1)i^

     

  • μ02πIr(π+1)i^

     

(1)

B0=BAB+BDE+BBCD

=μ0i4πri^+μ0i4πri^-μ0i2ri^

=μ0i2πri^-μ0i2ri^

=μ0i2πr(1-π)i^

=-μ0i2πr(π-1)i^



Q 19 :

The magnetic field at the centre of a current carrying circular loop of radius R is 16μT. The magnetic field at a distance x=3R on its axis from the centre is______μT[2026]

  • 2

     

  • 22

     

  • 4

     

  • 8

     

(1)

μ0I2R=16μT

μ0IR22(x2+R2)3/2=μ0IR22×8R3=2μT