JEE Advanced 2004

LEVELJEE Main

Animated Solution for Physics - Magnetic Effects of Current: An electron moving with a speed $u$ along the positive $x$ -axis at $y = 0$ enters a region of uniform magnetic field $B = - B_{0} \hat{k}$ which exists to the right of $y$ -axis. The electron exits from the region after sometime with the speed $v$ at coordinate $y$ , then

Q: Why doesn't the magnetic field change the speed of the electron?

The magnetic force is always perpendicular to the velocity vector. Because work is the dot product of force and displacement ($W = \mathbf{F} \cdot d\mathbf{s}$), and the angle between them is $90^\circ$, the work done is zero. Zero work means zero change in kinetic energy, so the speed remains constant.

Q: What if the particle was a proton instead of an electron?

A proton has a positive charge. The cross product $\mathbf{v} \times \mathbf{B}$ would still point in the $+y$ direction, and multiplying by a positive charge means the force would be upwards ($+y$ direction). The proton would curve upwards and exit at $y > 0$.

Q: How do we know the path is a semicircle?

A charged particle moving perpendicular to a uniform magnetic field always moves in a circle. Since the field only exists for $x > 0$ and the particle enters at $x=0$ moving horizontally, it will complete exactly half of that circle before hitting the boundary at $x=0$ again.

Q: What would happen if the magnetic field was parallel to the velocity?

If $\mathbf{v}$ and $\mathbf{B}$ were parallel, the cross product $\mathbf{v} \times \mathbf{B}$ would be zero. The magnetic force would be zero, and the electron would continue moving in a straight line along the $x$-axis without any deflection.

Q: Does the mass of the electron affect the exit coordinate?

Yes! The radius of the circular path is given by $R = \frac{mv}{qB}$. A larger mass would result in a larger radius, meaning the electron would exit further down the negative $y$-axis (at $y = -2R$). However, it would still exit at a negative $y$-coordinate regardless of its mass.

Select Answer:

Visualized Solution

$v = u$

$The magnetic force on a moving charge is given by F_{m} = q (v \times B)$
$Since F_{m} ⊥ v, the work done by the magnetic field is zero.$
$W = Δ K = 0 ⟹ Kinetic energy is constant.$
$Therefore, the speed of the electron remains unchanged: v = u$

$F_{m} = - e u B_{0} \hat{j}$

$Initial velocity, v = u \hat{i}$
$Magnetic field, B = - B_{0} \hat{k}$
$Charge of electron, q = - e$
$F_{m} = - e (u \hat{i} \times - B_{0} \hat{k}) = e u B_{0} (\hat{i} \times \hat{k})$
$F_{m} = e u B_{0} (- \hat{j}) = - e u B_{0} \hat{j}$

$y < 0$

$The force is directed downwards (along - y axis).$
$This force provides the centripetal acceleration, causing the electron to move in a circular path.$
$The electron will trace a clockwise semicircle in the region x > 0 .$
$It will exit the magnetic field by crossing the y -axis at a point where y < 0 .$

$Conclusion$

$From our analysis:$
$1. The speed remains constant: v = u$
$2. The exit coordinate is negative: y < 0$
$Therefore, the correct option is (d).$

00:00 / 00:00

The Sigma Insight: Motion of a Charge in Magnetic Fields

Solution Diagram

Analyzing the Setup Imagine an electron cruising along the positive $x$ -axis

It's moving freely with a speed

u

until it crosses the

y

-axis and enters a new region. In this region (

x > 0

), there is a uniform magnetic field pointing directly into the page, represented mathematically as

B = - B_{0} \hat{k}

We need to figure out two things when the electron eventually exits this magnetic field region: its final speed

v

and the sign of its

y

-coordinate.

The Speed of the Electron Let's tackle the speed first

One of the most fundamental properties of the magnetic force is that it always acts perpendicular to the velocity of the moving charge.

Because the force is perpendicular to the displacement at every instant, the work done by the magnetic field on the electron is exactly zero. According to the work-energy theorem, if no work is done on an object, its kinetic energy cannot change.

W = Δ K = 0

If the kinetic energy remains constant, the speed must also remain constant. Therefore, the electron will exit the magnetic field with the exact same speed it had when it entered.

v = u

The Trajectory and Exit Coordinate Now, let's determine where the electron goes

We use the Lorentz force equation to find the direction of the force acting on the electron the moment it enters the field:

F_{m} = q (v \times B)

Let's plug in what we know. The velocity is

v = u \hat{i}

, the magnetic field is

B = - B_{0} \hat{k}

, and the charge of an electron is

q = - e

F_{m} = - e (u \hat{i} \times - B_{0} \hat{k})

First, let's evaluate the cross product:

\hat{i} \times \hat{k} = - \hat{j}

F_{m} = - e (- u B_{0} (- \hat{j})) = - e (u B_{0} \hat{j}) = - e u B_{0} \hat{j}

The resulting force is in the

- \hat{j}

direction, which means it points downwards along the negative

y

-axis.

This downward force acts as a centripetal force, causing the electron to curve downwards into a circular path. Since the magnetic field only exists for

x > 0

, the electron will trace out a semicircle in the clockwise direction.

When it completes this semicircle, it will cross the

y

-axis again to exit the magnetic field. Because it was forced downwards from the origin (

y = 0

), it will inevitably exit at a point where the

y

-coordinate is negative.

y < 0

Final Conclusion Combining our two findings, we know that the speed remains unchanged ( $v = u$ ) and the exit point is below the $x$ -axis ( $y < 0$ )

This perfectly matches option (d).

Common Traps & FAQs

Why doesn't the magnetic field change the speed of the electron?

The magnetic force is always perpendicular to the velocity vector. Because work is the dot product of force and displacement (

W = F \cdot d s

), and the angle between them is

9 0^{\circ}

, the work done is zero. Zero work means zero change in kinetic energy, so the speed remains constant.

What if the particle was a proton instead of an electron?

A proton has a positive charge. The cross product

v \times B

would still point in the

+ y

direction, and multiplying by a positive charge means the force would be upwards (

+ y

direction). The proton would curve upwards and exit at

y > 0

How do we know the path is a semicircle?

A charged particle moving perpendicular to a uniform magnetic field always moves in a circle. Since the field only exists for

x > 0

and the particle enters at

x = 0

moving horizontally, it will complete exactly half of that circle before hitting the boundary at

x = 0

again.

What would happen if the magnetic field was parallel to the velocity?

v

and

B

were parallel, the cross product

v \times B

would be zero. The magnetic force would be zero, and the electron would continue moving in a straight line along the

x

-axis without any deflection.

Does the mass of the electron affect the exit coordinate?

Yes! The radius of the circular path is given by

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

. A larger mass would result in a larger radius, meaning the electron would exit further down the negative

y

-axis (at

y = - 2 R

). However, it would still exit at a negative

y

-coordinate regardless of its mass.

Similar Questions

JEE Advanced 2007

LEVELJEE Main

A magnetic field $B = B_{0} \hat{j}$ exists in the region $a < x < 2 a$ and $B = - B_{0} \hat{j}$ , in the region $2 a < x < 3 a$ , where $B_{0}$ is a positive constant. A positive point charge moving with a velocity $v = v_{0} \hat{i}$ , where $v_{0}$ is a positive constant, enters the magnetic field at $x = a$ . The trajectory of the charge in this region can be like

The region between $y = 0$ and $y = d$ contains a magnetic field $B = B \hat{k}$ . A particle of mass $m$ and charge $q$ enters the region with a velocity $v = v \hat{i}$ . If $d = \frac{m v}{2 q B}$ , then the acceleration of the charged particle at the point of its emergence at the other side is

(A)

\frac{q v B}{m} (\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j})

(B)

\frac{q v B}{m} (\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j})

(C)

\frac{q v B}{m} (\frac{- j ^ + i ^}{2})

(D)

\frac{q v B}{m} (\frac{i ^ + j ^}{2})

JEE Main 2020

LEVELJEE Main

An electron is moving along +x-direction with a velocity of $6 \times 1 0^{6} ms^{- 1}$ . It enters a region of uniform electric field of $300 V/cm$ pointing along +y-direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x-direction will be

(A)

3 \times 1 0^{- 4} T

, along + z-direction

(B)

5 \times 1 0^{- 3} T

, along − z-direction

(C)

5 \times 1 0^{- 3} T

, along + z-direction

(D)

3 \times 1 0^{- 4} T

, along − z-direction

LEVELJEE Main

A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ -direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ -direction, extending from $x = a$ to $x = b$ . The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

(A)

\frac{q b B}{m}

(B)

\frac{q ( b - a ) B}{m}

(C)

\frac{q a B}{m}

(D)

\frac{q ( b + a ) B}{m}

JEE Main 2007

LEVELJEE Advanced

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $E$ and $B$ with a velocity $v$ perpendicular to both $E$ and $B$ and comes out without any change in magnitude or direction of $v$ . Then,

(A)

v = \frac{E \times B}{B ^{2}}

(B)

v = \frac{B \times E}{B ^{2}}

(C)

v = \frac{E \times B}{E ^{2}}

(D)

v = \frac{B \times E}{E ^{2}}

JEE Advanced 2017

LEVELJEE Advanced

Comprehension Passage

A charged particle (electron or proton) is introduced at the origin (

x = 0, y = 0, z = 0

) with a given initial velocity

v

. A uniform electric field

E

and a uniform magnetic field

B

exist everywhere. The velocity

v

, electric field

E

and magnetic field

B

are given in columns 1, 2 and 3, respectively. The quantities

E_{0}, B_{0}

are positive in magnitude. $\begin{array}{lll} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline \text{(I) Electron with } \mathbf{v} = 2\frac{E_0}{B_0}\hat{x} & \text{(i) } \mathbf{E} = E_0\hat{z} & \text{(P) } \mathbf{B} = -B_0\hat{x} \\ \text{(II) Electron with } \mathbf{v} = \frac{E_0}{B_0}\hat{y} & \text{(ii) } \mathbf{E} = -E_0\hat{y} & \text{(Q) } \mathbf{B} = B_0\hat{x} \\ \text{(III) Proton with } \mathbf{v} = 0 & \text{(iii) } \mathbf{E} = -E_0\hat{x} & \text{(R) } \mathbf{B} = B_0\hat{y} \\ \text{(IV) Proton with } \mathbf{v} = 2\frac{E_0}{B_0}\hat{x} & \text{(iv) } \mathbf{E} = E_0\hat{x} & \text{(S) } \mathbf{B} = B_0\hat{z} \\ \hline \end{array}$

Question 1:

In which case would the particle move in a straight line along the negative direction of Y-axis (i.e. move along $- \overset{y}{^}$ )?

In which case will the particle move in a straight line with constant velocity?

In which case will the particle describe a helical path with axis along the positive z-direction?

A uniform magnetic field $B$ exists in the region between $x = 0$ and $x = \frac{3 R}{2}$ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge $+ Q$ and momentum $p$ directed along $X$ -axis enters region 2 from region 1 at point $P_{1} (y = - R)$ . Which of the following option(s) is/are correct?

* Multiple Correct Options

(A)

When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point

P_{1}

and the farthest point from

Y

-axis is

\frac{p}{2}

(B)

For

B = \frac{8}{13} \frac{p}{QR}

, the particle will enter region 3 through the point

P_{2}

X

-axis.

(C)

For

B > \frac{2}{3} \frac{p}{QR}

, the particle will re-enter region 1.

(D)

For a fixed

B

, particles of same charge

Q

and same velocity

v

, the distance between the point

P_{1}

and the point of re-entry into region 1 is inversely proportional to the mass of the particle.

LEVELJEE Main

An electric charge $+ q$ moves with velocity $v = 3 \hat{i} + 4 \hat{j} + \hat{k}$ , in an electromagnetic field given by $E = 3 \hat{i} + \hat{j} + 2 \hat{k}$ , $B = \hat{i} + \hat{j} - 3 \hat{k}$ . The $y$ component of the force experienced by $+ q$ is

(A)

2 q

(B)

11 q

(C)

5 q

(D)

3 q

LEVELJEE Main

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then

(A)

its velocity will decrease

(B)

its velocity will increase

(C)

it will turn towards right of direction of motion

(D)

it will turn towards left of direction of motion

JEE Main 2003

LEVELJEE Main

A particle of charge $- 16 \times 1 0^{- 18}$ C moving with velocity $10 ms^{- 1}$ along the X-axis enters a region where a magnetic field of induction $B$ is along the Y-axis and an electric field of magnitude $1 0^{4} Vm^{- 1}$ is along the negative Z-axis. If the charged particle continues moving along the X-axis, the magnitude of $B$ is

(A)

1 0^{3} Wb/m^{2}

(B)

1 0^{5} Wb/m^{2}

(C)

1 0^{16} Wb/m^{2}

(D)

1 0^{- 3} Wb/m^{2}

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=u

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fm​=−euB0​j^​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y<0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup Imagine an electron cruising along the positive .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis

The Speed of the Electron Let's tackle the speed first

The Trajectory and Exit Coordinate Now, let's determine where the electron goes

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In which case would the particle move in a straight line along the negative direction of Y-axis (i.e. move along −y^​)?

In which case will the particle move in a straight line with constant velocity?

In which case will the particle describe a helical path with axis along the positive z-direction?

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then

$v = u$

$F_{m} = - e u B_{0} \hat{j}$

$y < 0$

$Conclusion$

Analyzing the Setup Imagine an electron cruising along the positive $x$ -axis

In which case would the particle move in a straight line along the negative direction of Y-axis (i.e. move along $- \overset{y}{^}$ )?