Q 1 :

An element l=xi^ is placed at the origin and carries a large current I = 10 A. The magnetic field on the y-axis at a distance of 0.5 m from the element x of 1 cm length is ______.                [2024]

  • 8×10-8 T

     

  • 10×10-8 T

     

  • 4×10-8 T

     

  • 12×10-8 T

     

(3)

By Biot-Savart law, dB=μ0i4π(dl×r)r3  (Tesla)

Since element is very small,

Given: dl=1100i^m,  r=12j^m,  i=10 A,  θ=90°

=10-7×10×(1100i^×12j^)(12)3

=10-7×10×8200k^=4×10-8 T(+k^)



Q 2 :

Two parallel long current-carrying wire separated by a distance 2r are shown in the figure. The ratio of magnetic field at A to the magnetic field produced at C is x7. The value of x is ______.             [2024]



(5)

BA=μ0I2πr+μ0(2I)2π(3r)=5μ0I6πr

BC=μ0(2I)2πr+μ0I2π(3r)=7μ0I6πr

  BABC=57x=5



Q 3 :

Two long, straight wires carry equal currents in opposite directions as shown in the figure. The separation between the wires is 5.0 cm. The magnitude of the magnetic field at a point P midway between the wires is ______ μT.

(Given: μ0=4π×10-7 TmA-1)                  [2024]



(160)

B=(μ0i2πa)×2=4π×10-7×10π×(52×10-2)

=16×10-5=160μT

 



Q 4 :

The current of 5A flows in a square loop of sides 1m is placed in air. The magnetic field at the centre of the loop is X2×10-7T. The value of X is ______.            [2024]



(40)

i = 5A,

Square of side = 1m

Magnetic field at centre will be into the plane (X), due to all four wires

Bnet=4B1

B1=Magnetic field due to one side,

B1=μ0.i4πr(sinθ1+sinθ2),  Here θ1=θ2=45°

r=12 m

B1=μ0.(5)4π(1/2)(12+12)

Bnet=4B1=402×10-2 tesla

 x=40



Q 5 :

A regular polygon of 6 sides is formed by bending a wire of length 4π meter. If an electric current of 4π3 A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x×10-7T. The value of x is ______.          [2024]



(72)

B=6(μ0I4πr)(sin30°+sin30°)

=610-7×4π3(3×4π2×6)=72×10-7 Tx=72

 



Q 6 :

Figure shows a current carrying square loop ABCD of edge length is 'a' lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is         [2025]

  • 3πμ0I2a

     

  • μ0I2πa

     

  • 2μ0I3πa

     

  • 2μ0I3πa

     

(3)

B=BAB+BBC+BCD+BDA

BAB=BBC

          =μ0(2I/3)4π(a/2)[sin45°+sin45°](k^)

          =μ0I23πa(k^)

BCD=BDA

           =μ0(I/3)4π(a/2)[sin45°+sin45°](k^)

          =μ0I26πa(k^)

B=2[μ0I23πa(k^)+μ0I26πa(k^)]

 B=2μ0I3πa(k^)



Q 7 :

Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is x×105 T. The value of x is _____. Take permeability of free space as μ0=4π×107 SI units.          [2025]



(1)

At P

B=B1+B2=μ0i22πr2k^μ0i12πr1k^

B=μ02π(44×102510×102)k^

     =1×105 T

 x=1



Q 8 :

A current of 5 A exists in a square loop of side 12m. Then the magnitude of the magnetic field B at the centre of the square loop will be p×106 T. Where, value of p is ______. [Take μ0=4π×107 T mA1].         [2025]



(8)

Let B be the magnetic field due to single side then

B=μ0i4πd(sin θ1+sin θ2)

B=107×5×2122×12=2×106

 Bnet at centre O = 4 B

Bnet=8×106 T

 p=8



Q 9 :

A circular loop of radius r is carrying current 1 A. The ratio of magnetic field at the center of the circular loop and at a distance r from the center of the loop on its axis is     [2023]
 

  • 1:32

     

  • 22:1

     

  • 32:2

     

  • 1:2

     

(2)

Magnetic field due to a current carrying circular loop on its axis is given as

B=μ0ir22(r2+x2)3/2

At centre, x=0, B1=μ0i2r

At x=r, B2=μ0i2×22r

B1B2=22



Q 10 :

The magnitude of magnetic induction at mid-point O due to current arrangement as shown in the figure will be                      [2023]

  • μ0I2πa

     

  • 0

     

  • μ0I4πa

     

  • μ0Iπa

     

(4)

Magnetic field due to current in BC and ET are outward at point 'O'

B0=μ0i4πr+μ0i4πr=μ0i2πr=μ0iπa



Q 11 :

As shown in the figure, a current of 2 A flowing in an equilateral triangle of side 43cm. The magnetic field at the centroid O of the triangle is                 [2023]

(Neglect the effect of earth's magnetic field.)

  • 43×10-4 T

     

  • 43×10-5 T

     

  • 3×10-4 T

     

  • 33×10-5 T

     

(4)

d tan 60°=23

d=2 cm

B=3×μ0i2πdsin60°

=3×2×10-7×22×10-2×32

=33×10-5



Q 12 :

Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is ________ ×10-6 T. (Given: 2=1.4)     [2023]



(68)

Magnetic fields due to both wires will be perpendicular to each other.

B1=μ0i12πd,  B2=μ0i22πd

Bnet=B12+B22=μ02πdi12+i22

4π×10-72π×(72)×10-2×82+152 (d=72 cm)

=68×10-6 T



Q 13 :

An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76×106 ms-1 in an orbit of radius 0.52Å. The magnetic field produced at the nucleus of the hydrogen atom is _________ T.                       [2023]



(40)

Magnetic field due to moving charge

B=μ04πqvsinθr2

B=μ04πevsin(π/2)r2

B=10-7×1.6×10-19×6.76×1060.52×0.52×10-20

  B=40 T



Q 14 :

A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is :

A. maximum at either ends of the conductor and minimum at the midpoint
B. maximum at the axis of the conductor
C. minimum at the surface of the conductor
D. minimum at the axis of the conductor
E. same at all points in the cross-section of the conductor

Choose the correct answer from the options given below :   [2026]

  • B, C Only

     

  • D Only

     

  • A, D Only

     

  • E Only

     

(2)

Solid cylinder

Bmax at surface

Bmin at axis



Q 15 :

The current passing through a conducting loop in the form of equilateral triangle of side 43 cm is 2 A. The magnetic field at its centroid is α×10-5 T The value of α is ______.

(Given: μ0=4π×10-7 SI units)  [2026]

  • 32

     

  • 3

     

  • 33

     

  • 23

     

(3)

B=μ04π×Id[sin60°+sin60°]×3

B=10-7×22×10-2(32+32)×3

=3×10-5×3=33×10-5