JEE Main Magnetic Effect of Current PYQ

6. (JEE Main 2024 (Online) 8th April Morning Shift )

An electron with kinetic energy $5 eV$ enters a region of uniform magnetic field of 3 $μ T$ 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 electron moves along the same path, is __________ ${NC}^{- 1}$ .

(Given, mass of electron $= 9 \times 10^{- 31} kg$ , electric charge $= 1.6 \times 10^{- 19} C$ )

Correct answer is 4

An electron with kinetic energy $5 eV$ enters a region of uniform magnetic field of 3 $μ T$ 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 electron moves along the same path, is __________ ${NC}^{- 1}$ .

(Given, mass of electron $= 9 \times 10^{- 31} kg$ , electric charge $= 1.6 \times 10^{- 19} C$ )

7. (JEE Main 2024 (Online) 31st January Morning Shift )

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 _________ $T$ .

Correct answer is 5

$\begin{aligned} \vec{F} = q (\vec{v} \times \vec{B}) \\ 5 e \hat{k} = e (3 \hat{i} + 5 \hat{j}) \times (B_{0} \hat{i} + 2 B_{0} \hat{j}) \\ 5 e \hat{k} = e (6 B_{0} \hat{k} - 5 B_{0} \hat{k}) \\ \Rightarrow B_{0} = 5 T \end{aligned}$

8. (JEE Main 2024 (Online) 29th January Evening Shift )

A charge of $4.0 μ C$ is moving with a velocity of $4.0 \times 10^{6} {ms}^{- 1}$ along the positive $y$ axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) T$ . The force acting on the charge is $x \hat{i} N$ . The value of $x$ is __________.

Correct answer is 32

$\begin{aligned} q & = 4 μ C, \vec{v} = 4 \times 10^{6} \hat{j} m / s \\ \vec{B} & = 2 \hat{k T} \\ \vec{F} & = q (\vec{v} \times \vec{B}) \\ = 4 \times 10^{- 6} (4 \times 10^{6} \hat{j} \times 2 \hat{k}) \\ = 4 \times 10^{- 6} \times 8 \times 10^{6} \hat{i} \\ \vec{F} & = 32 \hat{i} N \\ x & = 32 \end{aligned}$

9. (JEE Main 2023 (Online) 13th April Evening Shift )

An electron is moving along the positive $x$ -axis. If the uniform magnetic field is applied parallel to the negative z-axis, then

A. The electron will experience magnetic force along positive y-axis

B. The electron will experience magnetic force along negative y-axis

C. The electron will not experience any force in magnetic field

D. The electron will continue to move along the positive $x$ -axis

E. The electron will move along circular path in magnetic field

Choose the correct answer from the options given below:

A. A and E only

B. B and D only

C. B and E only

D. C and D only

Correct Answer is Option (C)

The Lorentz force equation is given as:

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

The electron is moving along the positive x-axis, so its velocity vector is $\vec{v} = v_{x} \hat{i}$ . The magnetic field is applied parallel to the negative z-axis, so its magnetic field vector is $\vec{B} = - B_{z} \hat{k}$ .

Now, we can calculate the cross product of the velocity and magnetic field vectors:

$\vec{v} \times \vec{B} = (v_{x} \hat{i}) \times (- B_{z} \hat{k})$

Using the cross product properties, we get:

$\vec{v} \times \vec{B} = - v_{x} B_{z} (\hat{i} \times \hat{k})$

The cross product of $\hat{i}$ and $\hat{k}$ is $- \hat{j}$ , so:

$\vec{v} \times \vec{B} = - v_{x} B_{z} (- \hat{j}) = v_{x} B_{z} \hat{j}$

Since the electron has a negative charge, the magnetic force will be in the opposite direction:

$\vec{F} = - (- e) (v_{x} B_{z} \hat{j}) = e (v_{x} B_{z} \hat{j})$

As a result, the electron will experience a magnetic force along the negative y-axis.

Additionally, as mentioned earlier, when a charged particle moves through a magnetic field perpendicular to its velocity, it follows a circular path. In this case, the velocity of the electron is along the positive x-axis, and the magnetic field is along the negative z-axis, which are indeed perpendicular to each other. As a result, the electron will move along a circular path in the magnetic field.

Hence, the correct answer is:

(C) B and E only

10. (JEE Main 2023 (Online) 11th April Evening Shift )

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

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:

A. B, C and D only

B. B and C only

C. A and D only

D. B and E only

Correct Answer is Option (B)

The magnetic field inside a solenoid is uniform and parallel to the axis of the solenoid. When an electron moves with constant velocity along the axis of the solenoid, the angle between its velocity vector and the magnetic field is 0°.

The magnetic force experienced by a moving charge is given by the Lorentz force formula:

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

Since the angle between the velocity vector and the magnetic field is 0°, the cross product term becomes zero:

$\vec{v} \times \vec{B} = 0$

Therefore, the magnetic force experienced by the electron is also zero:

$\vec{F} = 0$

As a result, the electron will not experience any magnetic force (Option B) and will continue to move along the axis of the solenoid with constant velocity (Option C).

Thus, the correct answer is:

B and C only