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

Q 1 :

The electrostatic force $({\vec{F}}_{1})$ and magnetic force $({\vec{F}}_{2})$ acting on a charge $q$ moving with velocity $v$ can be written [2024]

${\vec{F}}_{1} = q \vec{B}, {\vec{F}}_{2} = q (\vec{B} \times \vec{v})$
${\vec{F}}_{1} = q \vec{E}, {\vec{F}}_{2} = q (\vec{v} \times \vec{B})$
${\vec{F}}_{1} = q \vec{v} \cdot \vec{E}, {\vec{F}}_{2} = q (\vec{B} \cdot \vec{v})$
${\vec{F}}_{1} = q \vec{E}, {\vec{F}}_{2} = q (\vec{B} \times \vec{v})$

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(2)

$The electrostatic force acting on a charge, {\vec{F}}_{1} = q \vec{E}$

$The magnetic force acting on a charge {\vec{F}}_{2} = q (\vec{v} \times \vec{B})$

Q 2 :

A proton and a deuteron $(q = + e, m = 2.0 u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$ , moving perpendicular to $\vec{B}$ . The ratio of the radius $r_{d}$ of deuteron path to the radius $r_{p}$ of the proton path is: [2024]

$1 : 1$
$1 : \sqrt{2}$
$\sqrt{2} : 1$
$1 : 2$

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(3)

$As radius of circular path (r) = \frac{m v}{q B}$ $\dots (i)$

$Given {(K E)}_{p} = {(K E)}_{d}$

$\frac{1}{2} m_{p} v_{p}^{2} = \frac{1}{2} m_{d} v_{d}^{2} \Rightarrow m_{p} v_{p}^{2} = m_{d} v_{d}^{2}$

$As m_{d} = 2 m_{p} \Rightarrow m_{p} v_{p}^{2} = 2 m_{p} v_{d}^{2} \Rightarrow v_{p} = \sqrt{2} v_{d}$

Now from equation (i),

$\Rightarrow \frac{r_{p}}{r_{d}} = \frac{m_{p} v_{p}}{e B} \times \frac{e B}{2 m_{p} v_{d}} = \frac{1}{\sqrt{2}} = r_{d} : r_{p} = \sqrt{2} : 1$

Q 3 :

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If $\bar{E}$ and $\bar{B}$ represent the electric and magnetic fields respectively, then the region of space may have:

(A) E = 0, B = 0 (B) E = 0, B ≠ 0

(C) E ≠ 0, B = 0 (D) E ≠ 0, B ≠ 0

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

(A), (B) and (C) only
(A), (C) and (D) only
(A), (B) and (D) only
(B), (C) and (D) only

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(3)

Net force on particle must be zero i.e.

$q \vec{E} + q \vec{V} \times \vec{B} = 0$

Possible cases are

(i) $\vec{E} and \vec{B} = 0$

(ii) $\vec{V} \times \vec{B} = 0, \vec{E} = 0$

(iii) $q \vec{E} = - q \vec{V} \times \vec{B}$

$\vec{E} \neq 0 and \vec{B} \neq 0$

Q 4 :

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_{1}$ and $R_{2}$ respectively. The mass ratio of X and Y is [2024]

$(\frac{R_{1}}{R_{2}})$
${(\frac{R_{2}}{R_{1}})}^{2}$
$(\frac{R_{2}}{R_{1}})$
${(\frac{R_{1}}{R_{2}})}^{2}$

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(4)

$R = \frac{m v}{q B} = \frac{p}{q B} = \frac{\sqrt{2 m (KE)}}{q B} = \frac{\sqrt{2 m q V}}{q B}$

$R \propto \sqrt{m} \Rightarrow m \propto R^{2}$

$\frac{m_{1}}{m_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$

Q 5 :

An electron with kinetic energy 5 eV enters a region of uniform magnetic field of 3 μT perpendicular to its direction. An electric field E is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that the electron moves along the same path, is ____ $N C^{- 1}$ . (Given, mass of electron $= 9 \times 10^{- 31}$ kg, electric charge $= 1.6 \times 10^{- 19}$ C) [2024]

(4) For the given condition of moving undeflected, net force should be zero.

$q E = q V B \Rightarrow E = V B$

$E = \sqrt{\frac{2 \times K E}{m}} \times B = \sqrt{\frac{2 \times 5 \times 1.6 \times 10^{- 19}}{9 \times 10^{- 31}}} \times 3 \times 10^{- 6} = 4 N / C$

Q 6 :

An electron moves through a uniform magnetic field $\vec{B} = B_{0} \hat{i} + 2 B_{0} \hat{J} T$ . At a particular instant of time, the velocity of electron is $\vec{u} = 3 \hat{i} + 5 \hat{j}$ m/s. If the magnetic force acting on electron is $\vec{F} = 5 e \hat{k}$ $N$ , where e is the charge of electron, then the value of $B_{0}$ is ____ T. [2024]

(5) $\vec{F} = q (\vec{v} \times \vec{B})$

$5 e \hat{k} = e (3 \hat{i} + 5 \hat{j}) \times (B_{0} \hat{i} + 2 B_{0} \hat{j})$

$5 e \hat{k} = e (6 B_{0} \hat{k} - 5 B_{0} \hat{k}) \Rightarrow B_{0} = 5 T$

Q 7 :

An electron is projected with uniform velocity along the axis inside a current-carrying long solenoid. Then [2024]

the electron will continue to move with uniform velocity along the axis of the solenoid
the electron will be accelerated along the axis
the electron path will be circular about the axis
the electron will experience a force at 45° to the axis and execute a helical path

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(1)

The force on a charged particle moving in a magnetic field, $\vec{F} = q (\vec{v} \times \vec{B})$

Since $\vec{v} ∥ \vec{B}$ so the force on the electron due to the magnetic field is zero.

So it will move along the axis with uniform velocity.

Q 8 :

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.

Reason (R) : The magnetic field in that region is along the direction of velocity of the electron.

In the light of the above statements, choose the correct answer from the options given below: [2025]

(A) is false but (R) is true
Both (A) and (R) are true and (R) is the correct explanation of (A)
Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(A) is true but (R) is false

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(2)

$\vec{F} = q (\vec{v} \times \vec{B})$

$\vec{F} = 0 when \vec{v} ∥ \vec{B}$

In this case, $θ = 0 ° or 180 °$

Q 9 :

Consider a long thin conducting wire carrying a uniform current $I$ . A particle having mass "M" and charge "q' is released at a distance "a" from the wire with a speed $v_{0}$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [ $μ_{0}$ is vacuum permeability] [2025]

$a [1 - \frac{m v_{0}}{2 q μ_{0} I}]$
$\frac{a}{2}$
$a [1 - \frac{m v_{α}}{q μ_{0} I}]$
$a e^{\frac{- 4 π m v_{0}}{q μ_{0} I}}$

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(4)

Let the velocity of the particle make an angle $θ$ with initial direction when it is at a distance x.

$\Rightarrow \frac{d x}{d t} = - v_{0} \sin θ$ ... (i)

Also, $\frac{d θ}{d t} = \frac{q}{m} (\frac{μ_{0} I}{2 π x})$ ... (ii)

$\Rightarrow \frac{d x}{d θ} = \frac{- 2 π m v_{0} x \sin θ}{q μ_{0} I}$ ... dividing eq (i) with eq (ii)

or $\int_{a}^{x} \frac{d x}{x} = \frac{- 2 π m v_{0}}{q μ_{0} I} \int_{0}^{π} \sin θ d θ$

$\ln \frac{x}{a} = \frac{- 4 π m v_{0}}{q μ_{0} I}$

Or $x = a e^{- \frac{4 π m v_{0}}{μ_{0} I q}}$

Q 10 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : If oxygen ion ( $O^{- 2}$ ) and Hydrogen ion ( $H^{+}$ ) enter normal to the magnetic field with equal momentum, then the path of $O^{- 2}$ ion has a smaller curvature than that of $H^{+}$ .

Reason R : A proton with same linear momentum as an electron will from a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.

In the light of the above statement, choose the correct answer from the options given below: [2025]

A is true but R is false
Both A and R true but R is NOT the correct explanation of A
A is false but R is true
Both A and R true and R is the correct explanation of A

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(1)

Radius in magnetic field, $R = \frac{m v}{q B}$

(A) mv, B are same for $O^{2 -}$ and $H^{+}$ therefore $R \propto \frac{1}{| q |}$ therefore (A) is correct

(B) $R = \frac{m v}{q B}$ ; mv, q, B are same, therefore (R) is incorrect.

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]

(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]

(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]

(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]

(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]

(40)

$B = \frac{π}{2} \times 10^{- 3}$

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

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

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

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

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

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

Q 20 :

A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy : [2026]

a > g
a = g
0 < a < g
a = 0

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(2)

Since the solenoid is placed vertically, the magnetic field inside the solenoid will be either along $- y$ or $+ y$ axis.

$\Rightarrow Particle will gain velocity along - y axis.$

$\Rightarrow {\vec{F}}_{B} = q (\vec{v} \times \vec{B})$

$\Rightarrow {\vec{F}}_{B} = 0$

$\Rightarrow {\vec{F}}_{net} = m \vec{g}$

$\Rightarrow a_{net} = g$

Topic Question Set

Q 1 :

The electrostatic force $({\vec{F}}_{1})$ and magnetic force $({\vec{F}}_{2})$ acting on a charge $q$ moving with velocity $v$ can be written [2024]

Q 2 :

A proton and a deuteron $(q = + e, m = 2.0 u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$ , moving perpendicular to $\vec{B}$ . The ratio of the radius $r_{d}$ of deuteron path to the radius $r_{p}$ of the proton path is: [2024]

Q 4 :

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_{1}$ and $R_{2}$ respectively. The mass ratio of X and Y is [2024]

Q 7 :

An electron is projected with uniform velocity along the axis inside a current-carrying long solenoid. Then [2024]

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]

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]

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]

Q 20 :

A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy : [2026]

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Topic Question Set

Q 1 : The electrostatic force (F→1) and magnetic force (F→2) acting on a charge q moving with velocity v can be written [2024]

Q 2 : A proton and a deuteron (q=+e,m=2.0u) having same kinetic energies enter a region of uniform magnetic field B→ , moving perpendicular to B→. The ratio of the radius rd of deuteron path to the radius rp of the proton path is: [2024]

Q 4 : Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii R1 and R2 respectively. The mass ratio of X and Y is [2024]

Q 6 : An electron moves through a uniform magnetic field B→=B0i^+2B0J^T. At a particular instant of time, the velocity of electron is u→=3i^+5j^ m/s. If the magnetic force acting on electron is F→=5ek^ N, where e is the charge of electron, then the value of B0 is ____ T. [2024]

Q 7 : An electron is projected with uniform velocity along the axis inside a current-carrying long solenoid. Then [2024]

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]

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]

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]

Q 20 : A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy : [2026]

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Q 1 :

The electrostatic force $({\vec{F}}_{1})$ and magnetic force $({\vec{F}}_{2})$ acting on a charge $q$ moving with velocity $v$ can be written [2024]

Q 2 :

A proton and a deuteron $(q = + e, m = 2.0 u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$ , moving perpendicular to $\vec{B}$ . The ratio of the radius $r_{d}$ of deuteron path to the radius $r_{p}$ of the proton path is: [2024]

Q 4 :

Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_{1}$ and $R_{2}$ respectively. The mass ratio of X and Y is [2024]

Q 7 :

An electron is projected with uniform velocity along the axis inside a current-carrying long solenoid. Then [2024]

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]

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]

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]

Q 20 :

A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy : [2026]