Source And Effects Of Magnetic Field | Physics JEE previous year questions with Solutions

Q 11 :

Uniform magnetic fields of different strengths ( $B_{1}$ and $B_{2}$ ) 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?

(Consider the velocity of the particle to be normal to the magnetic field and $B_{2} > B_{1}$ ) [2025]

$\frac{m v}{q B_{1}} (1 - \frac{B_{2}}{B_{1}}) \times 2$
$\frac{m v}{q B_{1}} (1 - \frac{B_{1}}{B_{2}})$
$\frac{m v}{q B_{1}} (1 - \frac{B_{2}}{B_{1}})$
$\frac{m v}{q B_{1}} (1 - \frac{B_{1}}{B_{2}}) \times 2$

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(4)

As $\vec{v}$ is $⊥$ to $\vec{B}$ , so charge particle will move in circular path, whose radius is given by

$R = \frac{m v}{q B}$

Starting point $\to A$

Ending point $\to C$

$∴$ Net displacement = AC

AC = CD – AD

$A C = \frac{2 m v}{q B_{1}} - \frac{2 m v}{q B_{2}}$

$A C = \frac{2 m v}{q B_{1}} [1 - \frac{B_{1}}{B_{2}}]$

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]

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(4)

$\frac{1}{2} m v^{2} = E \Rightarrow {(m v)}^{2} = 2 m E$

Also, $r = \frac{m v}{q B} \Rightarrow r = \frac{\sqrt{2 m E}}{q B}$

So, $r \propto {(E)}^{\frac{1}{2}}$

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 $2 \times 10^{5} {ms}^{- 1}$ . When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is $x \times 10^{4} N / C$ . the value of x is ________. (Take the mass of the proton $= 1.6 \times 10^{- 27} kg$ .) [2025]

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(2)

For uniform speed : Bvq = Eq $\Rightarrow$ E = Bv

$r = \frac{v m}{B q} \Rightarrow B = \frac{m v}{r q}$

$E = (\frac{m v}{r q}) v = \frac{m v^{2}}{r q}$

$E = \frac{1.6 \times 10^{- 27} \times 4 \times 10^{10}}{2 \times 10^{- 2} \times 1.6 \times 10^{- 19}} = 2 \times 10^{4} N / C$

$\Rightarrow x = 2$

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 $m_{e} = 9 \times 10^{- 31} kg$ , charge of electron $| q_{e} |$ $= 1.6 \times 10^{- 19} C, μ_{0} = 4 π \times 10^{- 7} \frac{N}{A^{2}}, 1 ns = 10^{- 9} s$ ] [2025]

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(250)

Since time period of a revolving charge is $\frac{2 π m}{q B}$

Where B = magnetic field

Due to a solenoid B = $μ_{0} n I$

$\Rightarrow \frac{2 π m}{q T} = μ_{0} n I$

$\Rightarrow n = \frac{2 π m}{q T μ_{0} I}$

$= \frac{2 π \times 9 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 75 \times 10^{- 9} \times 4 π \times 10^{- 7} \times 1.5}$

$\Rightarrow n = \frac{9 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 75 \times 10^{- 9} \times 2 \times 10^{- 7} \times 1.5}$

$\Rightarrow n = \frac{9 \times 10^{4}}{360} = 250$ per meter

Q 15 :

A particle of charge 1.6 $μ$ C and mass 16 $μ$ g 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]

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(10)

Angle between $\vec{v}$ of charge and $\vec{B}$ is 90° motion will be uniform circular motion time period is given by

$T = \frac{2 π m}{q B} = \frac{2 π \times 16 \times 10^{- 9} kg}{1.6 \times 10^{- 6} \times 6.28}$ = 0.01 seconds

Q 16 :

An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid. [2023]

A. The electron will experience magnetic force along the axis of the solenoid.
B. The electron will not experience magnetic force.
C. The electron will continue to move along the axis of the solenoid.
D. The electron will be accelerated along the axis of the solenoid.
E. The electron will follow parabolic path inside the solenoid.

Choose the correct answer from the options given below:

B, C and D only
B and C only
A and D only
B and E only

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(2)

$\vec{F} = q (\vec{v} \times \vec{B})$ $as the angle between \vec{v} and \vec{B} is 0 °$

$\vec{F} = 0$

Q 17 :

A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be [2023]

straight along the direction of magnetic field B
helical path with the axis perpendicular to the direction of magnetic field B
helical path with the axis along magnetic field B
circular path

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(3)

Due to the component of velocity $(v_{1})$ along the magnetic field, no force will be experienced by the particle.

So $v_{1}$ remains unchanged.

But due to the component $v_{2}$ magnetic force acts towards the centre, i.e., moving it in a circular path. So the path is helical with the axis parallel to the magnetic field B.

Q 18 :

A charge particle of $2 μ C$ accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes a semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is ________ $\times 10^{- 18} kg$ . [2023]

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(144)

$r = \frac{m v}{q B} = \frac{\sqrt{2 k m}}{q B}$

$m = \frac{r^{2} q^{2} B^{2}}{2 k}$

$m = \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{- 3} \times 4 \times 10^{- 3} \times 10^{- 12}}{2 \times (100) \times 10^{- 6}}$

$= 144 \times 10^{- 18} kg$

Q 19 :

A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude $\frac{π}{2} \times 10^{- 3} T$ . The angle between the direction of magnetic field and velocity of proton is $60 °$ . The pitch of the helical path taken by the proton is _______cm. (Take, mass of proton $= 1.6 \times 10^{- 27} kg$ and charge on proton $= 1.6 \times 10^{- 19} C$ .) [2023]

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(40)

$B = \frac{π}{2} \times 10^{- 3}$

$K.E. = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{2 K.E.}{m}}$

$Pitch = v \cos 60 ° \times time period of one rotation$

$= v \cos 60 ° \times \frac{2 π m}{e B}$

$= \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{- 19}}{1.6 \times 10^{- 27}}} \times \cos 60 ° \times \frac{2 π \times 1.6 \times 10^{- 27}}{1.6 \times 10^{- 19} \times \frac{π}{2} \times 10^{- 3}}$

$= 2 \times 10^{4} \times \frac{1}{2} \times 4 \times 10^{- 5}$

$= 4 \times 10^{- 1} m = 40 cm$

Topic Question Set

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]

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Q 15 :

A particle of charge 1.6 $μ$ C and mass 16 $μ$ g 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]

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Q 17 :

A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be [2023]

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