Q 11 :    

Uniform magnetic fields of different strengths ( B1and B2) both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m and charge q, at the interface at an instant, moves into the region 2 with velocity v and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?

Figure

(Consider the velocity of the particle to be normal to the magnetic field and B2>B1)          [2025]

  • mvqB1(1B2B1)×2

     

  • mvqB1(1B1B2)

     

  • mvqB1(1B2B1)

     

  • mvqB1(1B1B2)×2

     

(4)

As v is  to B, so charge particle will move in circular path, whose radius is given by

R=mvqB

Starting point A

Ending point C

Figure

 Net displacement = AC

AC = CDAD

AC=2mvqB12mvqB2

AC=2mvqB1[1B1B2]



Q 12 :    

A particle of charge q, mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius (r). Which of the following curves represents the variation of r with E?          [2025]

  • Figure

     

  • Figure

     

  • Figure

     

  • Figure

     

(4)

Figure

12mv2=E  (mv)2=2mE

Also, r=mvqB  r=2mEqB

Figure

So, r(E)12

Graph should be like



Q 13 :    

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×105 ms1. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is x×104 N/C. the value of x is ________. (Take the mass of the proton =1.6×1027 kg.)         [2025]



2

For uniform speed : Bvq = Eq  E = Bv

r=mvBq  B=mvrq

E=(mvrq)v=mv2rq

E=1.6×1027×4×10102×102×1.6×1019=2×104 N/C

 x=2



Q 14 :    

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is ________.

[Take mass of electron me=9×1031 kg, charge of electron |qe|=1.6×1019C, μ0=4π×107NA2, 1 ns=109 s]          [2025]



250

Since time period of a revolving charge is 2πmqB

Where B = magnetic field

Due to a solenoid Bμ0nI

 2πmqT=μ0nI

 n=2πmqTμ0I

            =2π×9×10311.6×1019×75×109×2×107×1.5

 n=9×10311.6×1019×75×109×2×107×1.5

 n=9×104360=250 per meter



Q 15 :    

A particle of charge 1.6 μC and mass 16 μg is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is __________ ms. (π = 3.14)          [2025]



10

Figure

Angle between v of charge and B is 90° motion will be uniform circular motion time period is given by

T=2πmqB=2π×16×109 kg1.6×106×6.28 = 0.01 seconds