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