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