Q 1 :    

Paramagnetic substances

A. align themselves along the directions of external magnetic field.

B. attract strongly towards external magnetic field.

C. has susceptibility little more than zero.

D. move from a region of strong magnetic field to weak magnetic field.

Choose the most appropriate answer from the options given below:                           [2024]

  • A, C only

     

  • B, D only

     

  • A, B, C, D

     

  • A, B, C only

     

(1)    

         (A) align themselves along the directions of external magnetic field.

         (C) has susceptibility little more than zero.

          A, C only

 



Q 2 :    

The coercivity of a magnet is 5×103 A/m. The amount of current required to be passed in a solenoid of length 30 cm and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is _____ A.                              [2024]



(10)            Length of solenoid (λ)=30cm, No. of turns(n)=150 and H=5×103 A/m

                  We know n=NII=Nn

                  I=5×103×0.3150=10A

 



Q 3 :    

Match List-I with List-II                              [2024]

Choose the correct answer from the options given below:

  • (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

     

  • (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

     

  • (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

     

  • (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

     

(1)

 

 



Q 4 :    

The relationship between the magnetic susceptibility (χ) and the magnetic permeability (μ) is given by:

(μ0 is the permeability of free space and μr is relative permeability)         [2025]

  • χ=μμ01

     

  • χ=μrμ0+1

     

  • χ=μT+1

     

  • χ=1μμ0

     

(1)

We have

μr=(1+χ)  χ=(μr1)

μ=μ0μr  μr=μμ0

  χ=(μμ01)



Q 5 :    

The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility χmg=1.2×105) is          [2025]

  • 65×103%

     

  • 56×105%

     

  • 56×104%

     

  • 53×105%

     

(1)

% change in B=BB%=BnewBoldBold×100%

                               =μniμ0niμ0ni×100%=(μμ0)μ0×100%

                               =(μ0μrμ0)μ0×100%

                                =(μr1)×100%

                                =χn×100%=1.2×103%