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Moving Charges & Magnetism

153725 To a charged particle which is moving with a constant initial velocity $\vec{v}$ , uniform magnetic field is applied in the direction of the velocity

1 the particle moves in a spiral path

2 the particle moves in a circular path

3 the particle moves in a parabolic path

4 there is no change in the motion of the particle

Explanation:

D Since, we know that magnetic force and particle moving a velocity $v$ in uniform magnetic field,
$F = qvB \sin θ$
According to question,
Field is applied in the direction of velocity,
Then, $θ = 0^{\circ}$
$F = qvB \sin θ$
$F = 0$
So, there is no change in the motion of the particle.

EAMCET-1992

Moving Charges & Magnetism

153726 An electron beam passes through a magnetic field of $2 \times 10^{- 3} Wb / m^{2}$ and an electric field of $1.0 \times 10^{4} V / m$ both acting simultaneously. The path of electron remains undeviated.
The speed of electron if the electric field is removed and the radius of electron path will be respectively.

1

10 \times 10^{6} m / s, 2.43 cm

2

2.5 \times 10^{6} m / s, 0.43 cm

3

5 \times 10^{6} m / s, 1.43 cm

4 none of these

Explanation:

C Given,
$E = 1.0 \times 10^{4} V / m$
$B = 2 \times 10^{- 3} Wb / m$
Speed of electron $(v) = \frac{E}{B}$
$v = \frac{1.0 \times 10^{4}}{2 \times 10^{- 3}} = 5 \times 10^{6} m / \sec$
Radius of electron path
$r = \frac{mv}{qB} = \frac{9.1 \times 10^{- 31} \times 5 \times 10^{6}}{1.6 \times 10^{- 19} \times 2 \times 10^{- 3}} = 0.014218 m$
$r ⊔ 1.43 cm .$

AIIMS-2011

Moving Charges & Magnetism

153728 When a proton is released from rest in a room, it starts with an initial acceleration $a_{0}$ towards West. When it is projected towards North with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards West. The electric and magnetic fields in the room are

1

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

up

2

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

down

3

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

up

4

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

down

Explanation:

B
Electric force acting on west direction
$F_{1} = {ma}_{0} = qE$
$F_{1} = \frac{{ma}_{0}}{q} = \frac{{ma}_{o}}{e} (West)$

$F_{1} + F_{2} = m 3 a_{0}$
${ma}_{0} + {qv}_{0} B = {m 3 a}_{0} (∵ F_{1} = {ma}_{0})$
${qv}_{0} B = 2 {ma}_{0}$
$B = \frac{2 {ma}_{0}}{{qv}_{0}} = \frac{2 {ma}_{0}}{{ev}_{0}} (downward)$
$∵ F_{2} = {qv}_{0} B$

AP EAMCET (Medical)-24.04.2019

Moving Charges & Magnetism

153730 If $E$ and $B$ are the magnitudes of electric and magnetic fields respectively in some region of space, then the possibilities for which a charged particle may move in that space with a uniform velocity of magnitude $v$ are

1

E = vB

2

E \neq 0, B = 0

3

E = 0, B \neq 0

4

E \neq 0, B \neq 0

Explanation:

A For uniform velocity, net force $(F) = 0$ According to the Lorentz force
And
$= q (\vec{E} + \vec{E} \times \vec{B})$
$\vec{E} = - \vec{v} \times \vec{B})$
$E = vB \sin θ$
$E = vB (\vec{v} ⊥ \vec{B}, θ = 90^{\circ})$

WB JEE 2013

Moving Charges & Magnetism

153725 To a charged particle which is moving with a constant initial velocity $\vec{v}$ , uniform magnetic field is applied in the direction of the velocity

1 the particle moves in a spiral path

2 the particle moves in a circular path

3 the particle moves in a parabolic path

4 there is no change in the motion of the particle

Explanation:

D Since, we know that magnetic force and particle moving a velocity $v$ in uniform magnetic field,
$F = qvB \sin θ$
According to question,
Field is applied in the direction of velocity,
Then, $θ = 0^{\circ}$
$F = qvB \sin θ$
$F = 0$
So, there is no change in the motion of the particle.

EAMCET-1992

Moving Charges & Magnetism

153726 An electron beam passes through a magnetic field of $2 \times 10^{- 3} Wb / m^{2}$ and an electric field of $1.0 \times 10^{4} V / m$ both acting simultaneously. The path of electron remains undeviated.
The speed of electron if the electric field is removed and the radius of electron path will be respectively.

1

10 \times 10^{6} m / s, 2.43 cm

2

2.5 \times 10^{6} m / s, 0.43 cm

3

5 \times 10^{6} m / s, 1.43 cm

4 none of these

Explanation:

C Given,
$E = 1.0 \times 10^{4} V / m$
$B = 2 \times 10^{- 3} Wb / m$
Speed of electron $(v) = \frac{E}{B}$
$v = \frac{1.0 \times 10^{4}}{2 \times 10^{- 3}} = 5 \times 10^{6} m / \sec$
Radius of electron path
$r = \frac{mv}{qB} = \frac{9.1 \times 10^{- 31} \times 5 \times 10^{6}}{1.6 \times 10^{- 19} \times 2 \times 10^{- 3}} = 0.014218 m$
$r ⊔ 1.43 cm .$

AIIMS-2011

Moving Charges & Magnetism

153728 When a proton is released from rest in a room, it starts with an initial acceleration $a_{0}$ towards West. When it is projected towards North with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards West. The electric and magnetic fields in the room are

1

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

up

2

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

down

3

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

up

4

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

down

Explanation:

B
Electric force acting on west direction
$F_{1} = {ma}_{0} = qE$
$F_{1} = \frac{{ma}_{0}}{q} = \frac{{ma}_{o}}{e} (West)$

$F_{1} + F_{2} = m 3 a_{0}$
${ma}_{0} + {qv}_{0} B = {m 3 a}_{0} (∵ F_{1} = {ma}_{0})$
${qv}_{0} B = 2 {ma}_{0}$
$B = \frac{2 {ma}_{0}}{{qv}_{0}} = \frac{2 {ma}_{0}}{{ev}_{0}} (downward)$
$∵ F_{2} = {qv}_{0} B$

AP EAMCET (Medical)-24.04.2019

Moving Charges & Magnetism

153730 If $E$ and $B$ are the magnitudes of electric and magnetic fields respectively in some region of space, then the possibilities for which a charged particle may move in that space with a uniform velocity of magnitude $v$ are

1

E = vB

2

E \neq 0, B = 0

3

E = 0, B \neq 0

4

E \neq 0, B \neq 0

Explanation:

A For uniform velocity, net force $(F) = 0$ According to the Lorentz force
And
$= q (\vec{E} + \vec{E} \times \vec{B})$
$\vec{E} = - \vec{v} \times \vec{B})$
$E = vB \sin θ$
$E = vB (\vec{v} ⊥ \vec{B}, θ = 90^{\circ})$

WB JEE 2013

Moving Charges & Magnetism

153725 To a charged particle which is moving with a constant initial velocity $\vec{v}$ , uniform magnetic field is applied in the direction of the velocity

1 the particle moves in a spiral path

2 the particle moves in a circular path

3 the particle moves in a parabolic path

4 there is no change in the motion of the particle

Explanation:

D Since, we know that magnetic force and particle moving a velocity $v$ in uniform magnetic field,
$F = qvB \sin θ$
According to question,
Field is applied in the direction of velocity,
Then, $θ = 0^{\circ}$
$F = qvB \sin θ$
$F = 0$
So, there is no change in the motion of the particle.

EAMCET-1992

Moving Charges & Magnetism

153726 An electron beam passes through a magnetic field of $2 \times 10^{- 3} Wb / m^{2}$ and an electric field of $1.0 \times 10^{4} V / m$ both acting simultaneously. The path of electron remains undeviated.
The speed of electron if the electric field is removed and the radius of electron path will be respectively.

1

10 \times 10^{6} m / s, 2.43 cm

2

2.5 \times 10^{6} m / s, 0.43 cm

3

5 \times 10^{6} m / s, 1.43 cm

4 none of these

Explanation:

C Given,
$E = 1.0 \times 10^{4} V / m$
$B = 2 \times 10^{- 3} Wb / m$
Speed of electron $(v) = \frac{E}{B}$
$v = \frac{1.0 \times 10^{4}}{2 \times 10^{- 3}} = 5 \times 10^{6} m / \sec$
Radius of electron path
$r = \frac{mv}{qB} = \frac{9.1 \times 10^{- 31} \times 5 \times 10^{6}}{1.6 \times 10^{- 19} \times 2 \times 10^{- 3}} = 0.014218 m$
$r ⊔ 1.43 cm .$

AIIMS-2011

Moving Charges & Magnetism

153728 When a proton is released from rest in a room, it starts with an initial acceleration $a_{0}$ towards West. When it is projected towards North with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards West. The electric and magnetic fields in the room are

1

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

up

2

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

down

3

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

up

4

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

down

Explanation:

B
Electric force acting on west direction
$F_{1} = {ma}_{0} = qE$
$F_{1} = \frac{{ma}_{0}}{q} = \frac{{ma}_{o}}{e} (West)$

$F_{1} + F_{2} = m 3 a_{0}$
${ma}_{0} + {qv}_{0} B = {m 3 a}_{0} (∵ F_{1} = {ma}_{0})$
${qv}_{0} B = 2 {ma}_{0}$
$B = \frac{2 {ma}_{0}}{{qv}_{0}} = \frac{2 {ma}_{0}}{{ev}_{0}} (downward)$
$∵ F_{2} = {qv}_{0} B$

AP EAMCET (Medical)-24.04.2019

Moving Charges & Magnetism

153730 If $E$ and $B$ are the magnitudes of electric and magnetic fields respectively in some region of space, then the possibilities for which a charged particle may move in that space with a uniform velocity of magnitude $v$ are

1

E = vB

2

E \neq 0, B = 0

3

E = 0, B \neq 0

4

E \neq 0, B \neq 0

Explanation:

A For uniform velocity, net force $(F) = 0$ According to the Lorentz force
And
$= q (\vec{E} + \vec{E} \times \vec{B})$
$\vec{E} = - \vec{v} \times \vec{B})$
$E = vB \sin θ$
$E = vB (\vec{v} ⊥ \vec{B}, θ = 90^{\circ})$

WB JEE 2013

Moving Charges & Magnetism

153725 To a charged particle which is moving with a constant initial velocity $\vec{v}$ , uniform magnetic field is applied in the direction of the velocity

1 the particle moves in a spiral path

2 the particle moves in a circular path

3 the particle moves in a parabolic path

4 there is no change in the motion of the particle

Explanation:

D Since, we know that magnetic force and particle moving a velocity $v$ in uniform magnetic field,
$F = qvB \sin θ$
According to question,
Field is applied in the direction of velocity,
Then, $θ = 0^{\circ}$
$F = qvB \sin θ$
$F = 0$
So, there is no change in the motion of the particle.

EAMCET-1992

Moving Charges & Magnetism

153726 An electron beam passes through a magnetic field of $2 \times 10^{- 3} Wb / m^{2}$ and an electric field of $1.0 \times 10^{4} V / m$ both acting simultaneously. The path of electron remains undeviated.
The speed of electron if the electric field is removed and the radius of electron path will be respectively.

1

10 \times 10^{6} m / s, 2.43 cm

2

2.5 \times 10^{6} m / s, 0.43 cm

3

5 \times 10^{6} m / s, 1.43 cm

4 none of these

Explanation:

C Given,
$E = 1.0 \times 10^{4} V / m$
$B = 2 \times 10^{- 3} Wb / m$
Speed of electron $(v) = \frac{E}{B}$
$v = \frac{1.0 \times 10^{4}}{2 \times 10^{- 3}} = 5 \times 10^{6} m / \sec$
Radius of electron path
$r = \frac{mv}{qB} = \frac{9.1 \times 10^{- 31} \times 5 \times 10^{6}}{1.6 \times 10^{- 19} \times 2 \times 10^{- 3}} = 0.014218 m$
$r ⊔ 1.43 cm .$

AIIMS-2011

Moving Charges & Magnetism

153728 When a proton is released from rest in a room, it starts with an initial acceleration $a_{0}$ towards West. When it is projected towards North with a speed $v_{0}$ it moves with an initial acceleration $3 a_{0}$ towards West. The electric and magnetic fields in the room are

1

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

up

2

\frac{{ma}_{0}}{e}

West,

\frac{2 {ma}_{0}}{{ev}_{0}}

down

3

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

up

4

\frac{{ma}_{0}}{e}

East,

\frac{3 {ma}_{0}}{{ev}_{0}}

down

Explanation:

B
Electric force acting on west direction
$F_{1} = {ma}_{0} = qE$
$F_{1} = \frac{{ma}_{0}}{q} = \frac{{ma}_{o}}{e} (West)$

$F_{1} + F_{2} = m 3 a_{0}$
${ma}_{0} + {qv}_{0} B = {m 3 a}_{0} (∵ F_{1} = {ma}_{0})$
${qv}_{0} B = 2 {ma}_{0}$
$B = \frac{2 {ma}_{0}}{{qv}_{0}} = \frac{2 {ma}_{0}}{{ev}_{0}} (downward)$
$∵ F_{2} = {qv}_{0} B$

AP EAMCET (Medical)-24.04.2019

Moving Charges & Magnetism

153730 If $E$ and $B$ are the magnitudes of electric and magnetic fields respectively in some region of space, then the possibilities for which a charged particle may move in that space with a uniform velocity of magnitude $v$ are

1

E = vB

2

E \neq 0, B = 0

3

E = 0, B \neq 0

4

E \neq 0, B \neq 0

Explanation:

A For uniform velocity, net force $(F) = 0$ According to the Lorentz force
And
$= q (\vec{E} + \vec{E} \times \vec{B})$
$\vec{E} = - \vec{v} \times \vec{B})$
$E = vB \sin θ$
$E = vB (\vec{v} ⊥ \vec{B}, θ = 90^{\circ})$

WB JEE 2013