QBManager

Moving Charges & Magnetism

153533 A charged particle is moving in a uniform magnetic field in a circular path of radius ' $R$ '. When the energy of the particle becomes three times the original, the new radius will be

1

R

2

3 R

3

\frac{R}{3}

4

\sqrt{3} R

Explanation:

D Given, (K.E.) $)_{1} = 3$ (K.E.)
Radius of the circular path when charged particle moves in uniform magnetic field.
We know that, radius of circular path,
$R = \frac{mv}{Bq}$
And momentum $(p) = m v$
and kinetic energy, (K.E.) $= \frac{1}{2} {mv}^{2} = \frac{p^{2}}{2 m}$
$p = \sqrt{2 m (K . E .)}$
The radius of charged particle,
$R = \frac{\sqrt{2 m (K.E.)}}{Bq}$
$R \propto \sqrt{(K.E.)}$
Therefore, $\frac{R}{R_{1}} = \sqrt{\frac{(K.E.)}{(K.E.)_{1}}} = \sqrt{\frac{1}{3}}$
$R_{1} = \sqrt{3} R$

MHT-CET 2020

Moving Charges & Magnetism

153534 An electron moves in a circular arc of radius 10 $m$ at a constant speed of $2 \times 10^{7} {ms}^{- 1}$ with its plane of motion normal to a magnetic flux density of $10^{- 5} T$ . What will be the value of specific charge of the electron?

1

2 \times 10^{4} C {kg}^{- 1}

2

2 \times 10^{5} C {kg}^{- 1}

3

5 \times 10^{6} C {kg}^{- 1}

4

2 \times 10^{11} C {kg}^{- 1}

Explanation:

D Given, $r = 10 m, v = 2 \times 10^{7} m / \sec, B = 10^{- 5} T$ Force due to circular motion-

$F = \frac{{mv}^{2}}{r}$
Force due to magnetic field-
$F_{B} = qvB$
On equating equation (i) and (ii), we get-
$F = F_{B}$
$\frac{{mv}^{2}}{r} = qvB$
$\frac{q}{m} = \frac{v}{rB}$
Putting these value equation (iii), we get-
$\frac{q}{m} = \frac{2 \times 10^{7}}{10 \times 10^{- 5}}$
$\frac{q}{m} = 2 \times 10^{11} C {kg}^{- 1}$

BITSAT-2020]

Moving Charges & Magnetism

153537 A proton enters into a magnetic field of induction $1.732 T$ , with a velocity of $10^{7} m / s$ at an angle $60^{\circ}$ to the field. The force acting on the proton is $(e = 1.6 \times 10^{- 19} C, \sin 60^{\circ} = \cos 30^{\circ} =$ $\frac{\sqrt{3}}{2}$ )

1

1.6 \times 10^{- 12} N

2

3.2 \times 10^{- 12} N

3

4.0 \times 10^{- 12} N

4

2.4 \times 10^{- 12} N

Explanation:

D Given:
$B = 1.732 T$
$v = 10^{7} m / \sec$
$θ = 60^{\circ}$
$e = 1.6 \times 10^{- 19} C$
We know that, force acting on proton
$F_{P} = e (\vec{v} \times \vec{B})$
$= evB \sin θ$
$= evB \sin 60^{\circ}$
$= 1.6 \times 10^{- 19} \times 10^{7} \times 1.732 \times \frac{\sqrt{3}}{2}$
$F_{P} = 2.4 \times 10^{- 12} N$

MHT-CET 2019

Moving Charges & Magnetism

153538 Two particles carrying equal charges move parallel to each other with the speed $150 km / s$ . If $F_{1}$ and $F_{2}$ are magnetic and electric forces between two charged particles then,
$\frac{| F_{1} |}{| F_{2} |} is (Let μ_{0} ε_{0} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2})$

1

1.0 \times 10^{- 6}

2

1.5 \times 10^{- 7}

3

3.0 \times 10^{- 6}

4

2.5 \times 10^{- 7}

Explanation:

D Given, $F_{1} =$ magnetic force $& F_{2} =$ electric force
$v_{1} = v_{2} = 150 km / \sec$
$= 1.5 \times 10^{5} m / \sec$
$μ_{o} ε_{o} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2}$
Electric force between two moving charge particle is
$F_{2} = \frac{μ_{o}}{4 π ε_{o}} \frac{q^{2}}{r^{2}} \dots$
Magnetic force between two moving charge particle is-
$F_{1} = \frac{μ_{o}}{4 π} \frac{q_{1} q_{2} V_{1} V_{2}}{r^{2}}$
$F_{1} = \frac{μ_{o}}{4 π} \frac{q^{2} V^{2}}{r^{2}}$ ...(ii) $(∵ q = q_{2} = q, V_{1} = V_{2} = V)$
Equation (i) is divide by equation (ii)-
$\frac{F_{1}}{F_{2}} = \frac{\frac{μ_{o}}{4 π} \frac{q^{2} v^{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \frac{q^{2}}{r^{2}}} = ε_{o} μ_{o} v^{2}$
$\frac{F_{1}}{F_{2}} = ε_{0} μ_{0} v^{2}$
$= \frac{1}{9 \times 10^{16}} \times {(1.5 \times 10^{5})}^{2}$
$\frac{F_{1}}{F_{2}} = 2.5 \times 10^{- 7}$

TS- EAMCET-06.05.2019

Moving Charges & Magnetism

153533 A charged particle is moving in a uniform magnetic field in a circular path of radius ' $R$ '. When the energy of the particle becomes three times the original, the new radius will be

1

R

2

3 R

3

\frac{R}{3}

4

\sqrt{3} R

Explanation:

D Given, (K.E.) $)_{1} = 3$ (K.E.)
Radius of the circular path when charged particle moves in uniform magnetic field.
We know that, radius of circular path,
$R = \frac{mv}{Bq}$
And momentum $(p) = m v$
and kinetic energy, (K.E.) $= \frac{1}{2} {mv}^{2} = \frac{p^{2}}{2 m}$
$p = \sqrt{2 m (K . E .)}$
The radius of charged particle,
$R = \frac{\sqrt{2 m (K.E.)}}{Bq}$
$R \propto \sqrt{(K.E.)}$
Therefore, $\frac{R}{R_{1}} = \sqrt{\frac{(K.E.)}{(K.E.)_{1}}} = \sqrt{\frac{1}{3}}$
$R_{1} = \sqrt{3} R$

MHT-CET 2020

Moving Charges & Magnetism

153534 An electron moves in a circular arc of radius 10 $m$ at a constant speed of $2 \times 10^{7} {ms}^{- 1}$ with its plane of motion normal to a magnetic flux density of $10^{- 5} T$ . What will be the value of specific charge of the electron?

1

2 \times 10^{4} C {kg}^{- 1}

2

2 \times 10^{5} C {kg}^{- 1}

3

5 \times 10^{6} C {kg}^{- 1}

4

2 \times 10^{11} C {kg}^{- 1}

Explanation:

D Given, $r = 10 m, v = 2 \times 10^{7} m / \sec, B = 10^{- 5} T$ Force due to circular motion-

$F = \frac{{mv}^{2}}{r}$
Force due to magnetic field-
$F_{B} = qvB$
On equating equation (i) and (ii), we get-
$F = F_{B}$
$\frac{{mv}^{2}}{r} = qvB$
$\frac{q}{m} = \frac{v}{rB}$
Putting these value equation (iii), we get-
$\frac{q}{m} = \frac{2 \times 10^{7}}{10 \times 10^{- 5}}$
$\frac{q}{m} = 2 \times 10^{11} C {kg}^{- 1}$

BITSAT-2020]

Moving Charges & Magnetism

153537 A proton enters into a magnetic field of induction $1.732 T$ , with a velocity of $10^{7} m / s$ at an angle $60^{\circ}$ to the field. The force acting on the proton is $(e = 1.6 \times 10^{- 19} C, \sin 60^{\circ} = \cos 30^{\circ} =$ $\frac{\sqrt{3}}{2}$ )

1

1.6 \times 10^{- 12} N

2

3.2 \times 10^{- 12} N

3

4.0 \times 10^{- 12} N

4

2.4 \times 10^{- 12} N

Explanation:

D Given:
$B = 1.732 T$
$v = 10^{7} m / \sec$
$θ = 60^{\circ}$
$e = 1.6 \times 10^{- 19} C$
We know that, force acting on proton
$F_{P} = e (\vec{v} \times \vec{B})$
$= evB \sin θ$
$= evB \sin 60^{\circ}$
$= 1.6 \times 10^{- 19} \times 10^{7} \times 1.732 \times \frac{\sqrt{3}}{2}$
$F_{P} = 2.4 \times 10^{- 12} N$

MHT-CET 2019

Moving Charges & Magnetism

153538 Two particles carrying equal charges move parallel to each other with the speed $150 km / s$ . If $F_{1}$ and $F_{2}$ are magnetic and electric forces between two charged particles then,
$\frac{| F_{1} |}{| F_{2} |} is (Let μ_{0} ε_{0} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2})$

1

1.0 \times 10^{- 6}

2

1.5 \times 10^{- 7}

3

3.0 \times 10^{- 6}

4

2.5 \times 10^{- 7}

Explanation:

D Given, $F_{1} =$ magnetic force $& F_{2} =$ electric force
$v_{1} = v_{2} = 150 km / \sec$
$= 1.5 \times 10^{5} m / \sec$
$μ_{o} ε_{o} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2}$
Electric force between two moving charge particle is
$F_{2} = \frac{μ_{o}}{4 π ε_{o}} \frac{q^{2}}{r^{2}} \dots$
Magnetic force between two moving charge particle is-
$F_{1} = \frac{μ_{o}}{4 π} \frac{q_{1} q_{2} V_{1} V_{2}}{r^{2}}$
$F_{1} = \frac{μ_{o}}{4 π} \frac{q^{2} V^{2}}{r^{2}}$ ...(ii) $(∵ q = q_{2} = q, V_{1} = V_{2} = V)$
Equation (i) is divide by equation (ii)-
$\frac{F_{1}}{F_{2}} = \frac{\frac{μ_{o}}{4 π} \frac{q^{2} v^{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \frac{q^{2}}{r^{2}}} = ε_{o} μ_{o} v^{2}$
$\frac{F_{1}}{F_{2}} = ε_{0} μ_{0} v^{2}$
$= \frac{1}{9 \times 10^{16}} \times {(1.5 \times 10^{5})}^{2}$
$\frac{F_{1}}{F_{2}} = 2.5 \times 10^{- 7}$

TS- EAMCET-06.05.2019

Moving Charges & Magnetism

153533 A charged particle is moving in a uniform magnetic field in a circular path of radius ' $R$ '. When the energy of the particle becomes three times the original, the new radius will be

1

R

2

3 R

3

\frac{R}{3}

4

\sqrt{3} R

Explanation:

D Given, (K.E.) $)_{1} = 3$ (K.E.)
Radius of the circular path when charged particle moves in uniform magnetic field.
We know that, radius of circular path,
$R = \frac{mv}{Bq}$
And momentum $(p) = m v$
and kinetic energy, (K.E.) $= \frac{1}{2} {mv}^{2} = \frac{p^{2}}{2 m}$
$p = \sqrt{2 m (K . E .)}$
The radius of charged particle,
$R = \frac{\sqrt{2 m (K.E.)}}{Bq}$
$R \propto \sqrt{(K.E.)}$
Therefore, $\frac{R}{R_{1}} = \sqrt{\frac{(K.E.)}{(K.E.)_{1}}} = \sqrt{\frac{1}{3}}$
$R_{1} = \sqrt{3} R$

MHT-CET 2020

Moving Charges & Magnetism

153534 An electron moves in a circular arc of radius 10 $m$ at a constant speed of $2 \times 10^{7} {ms}^{- 1}$ with its plane of motion normal to a magnetic flux density of $10^{- 5} T$ . What will be the value of specific charge of the electron?

1

2 \times 10^{4} C {kg}^{- 1}

2

2 \times 10^{5} C {kg}^{- 1}

3

5 \times 10^{6} C {kg}^{- 1}

4

2 \times 10^{11} C {kg}^{- 1}

Explanation:

D Given, $r = 10 m, v = 2 \times 10^{7} m / \sec, B = 10^{- 5} T$ Force due to circular motion-

$F = \frac{{mv}^{2}}{r}$
Force due to magnetic field-
$F_{B} = qvB$
On equating equation (i) and (ii), we get-
$F = F_{B}$
$\frac{{mv}^{2}}{r} = qvB$
$\frac{q}{m} = \frac{v}{rB}$
Putting these value equation (iii), we get-
$\frac{q}{m} = \frac{2 \times 10^{7}}{10 \times 10^{- 5}}$
$\frac{q}{m} = 2 \times 10^{11} C {kg}^{- 1}$

BITSAT-2020]

Moving Charges & Magnetism

153537 A proton enters into a magnetic field of induction $1.732 T$ , with a velocity of $10^{7} m / s$ at an angle $60^{\circ}$ to the field. The force acting on the proton is $(e = 1.6 \times 10^{- 19} C, \sin 60^{\circ} = \cos 30^{\circ} =$ $\frac{\sqrt{3}}{2}$ )

1

1.6 \times 10^{- 12} N

2

3.2 \times 10^{- 12} N

3

4.0 \times 10^{- 12} N

4

2.4 \times 10^{- 12} N

Explanation:

D Given:
$B = 1.732 T$
$v = 10^{7} m / \sec$
$θ = 60^{\circ}$
$e = 1.6 \times 10^{- 19} C$
We know that, force acting on proton
$F_{P} = e (\vec{v} \times \vec{B})$
$= evB \sin θ$
$= evB \sin 60^{\circ}$
$= 1.6 \times 10^{- 19} \times 10^{7} \times 1.732 \times \frac{\sqrt{3}}{2}$
$F_{P} = 2.4 \times 10^{- 12} N$

MHT-CET 2019

Moving Charges & Magnetism

153538 Two particles carrying equal charges move parallel to each other with the speed $150 km / s$ . If $F_{1}$ and $F_{2}$ are magnetic and electric forces between two charged particles then,
$\frac{| F_{1} |}{| F_{2} |} is (Let μ_{0} ε_{0} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2})$

1

1.0 \times 10^{- 6}

2

1.5 \times 10^{- 7}

3

3.0 \times 10^{- 6}

4

2.5 \times 10^{- 7}

Explanation:

D Given, $F_{1} =$ magnetic force $& F_{2} =$ electric force
$v_{1} = v_{2} = 150 km / \sec$
$= 1.5 \times 10^{5} m / \sec$
$μ_{o} ε_{o} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2}$
Electric force between two moving charge particle is
$F_{2} = \frac{μ_{o}}{4 π ε_{o}} \frac{q^{2}}{r^{2}} \dots$
Magnetic force between two moving charge particle is-
$F_{1} = \frac{μ_{o}}{4 π} \frac{q_{1} q_{2} V_{1} V_{2}}{r^{2}}$
$F_{1} = \frac{μ_{o}}{4 π} \frac{q^{2} V^{2}}{r^{2}}$ ...(ii) $(∵ q = q_{2} = q, V_{1} = V_{2} = V)$
Equation (i) is divide by equation (ii)-
$\frac{F_{1}}{F_{2}} = \frac{\frac{μ_{o}}{4 π} \frac{q^{2} v^{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \frac{q^{2}}{r^{2}}} = ε_{o} μ_{o} v^{2}$
$\frac{F_{1}}{F_{2}} = ε_{0} μ_{0} v^{2}$
$= \frac{1}{9 \times 10^{16}} \times {(1.5 \times 10^{5})}^{2}$
$\frac{F_{1}}{F_{2}} = 2.5 \times 10^{- 7}$

TS- EAMCET-06.05.2019

Moving Charges & Magnetism

153533 A charged particle is moving in a uniform magnetic field in a circular path of radius ' $R$ '. When the energy of the particle becomes three times the original, the new radius will be

1

R

2

3 R

3

\frac{R}{3}

4

\sqrt{3} R

Explanation:

D Given, (K.E.) $)_{1} = 3$ (K.E.)
Radius of the circular path when charged particle moves in uniform magnetic field.
We know that, radius of circular path,
$R = \frac{mv}{Bq}$
And momentum $(p) = m v$
and kinetic energy, (K.E.) $= \frac{1}{2} {mv}^{2} = \frac{p^{2}}{2 m}$
$p = \sqrt{2 m (K . E .)}$
The radius of charged particle,
$R = \frac{\sqrt{2 m (K.E.)}}{Bq}$
$R \propto \sqrt{(K.E.)}$
Therefore, $\frac{R}{R_{1}} = \sqrt{\frac{(K.E.)}{(K.E.)_{1}}} = \sqrt{\frac{1}{3}}$
$R_{1} = \sqrt{3} R$

MHT-CET 2020

Moving Charges & Magnetism

153534 An electron moves in a circular arc of radius 10 $m$ at a constant speed of $2 \times 10^{7} {ms}^{- 1}$ with its plane of motion normal to a magnetic flux density of $10^{- 5} T$ . What will be the value of specific charge of the electron?

1

2 \times 10^{4} C {kg}^{- 1}

2

2 \times 10^{5} C {kg}^{- 1}

3

5 \times 10^{6} C {kg}^{- 1}

4

2 \times 10^{11} C {kg}^{- 1}

Explanation:

D Given, $r = 10 m, v = 2 \times 10^{7} m / \sec, B = 10^{- 5} T$ Force due to circular motion-

$F = \frac{{mv}^{2}}{r}$
Force due to magnetic field-
$F_{B} = qvB$
On equating equation (i) and (ii), we get-
$F = F_{B}$
$\frac{{mv}^{2}}{r} = qvB$
$\frac{q}{m} = \frac{v}{rB}$
Putting these value equation (iii), we get-
$\frac{q}{m} = \frac{2 \times 10^{7}}{10 \times 10^{- 5}}$
$\frac{q}{m} = 2 \times 10^{11} C {kg}^{- 1}$

BITSAT-2020]

Moving Charges & Magnetism

153537 A proton enters into a magnetic field of induction $1.732 T$ , with a velocity of $10^{7} m / s$ at an angle $60^{\circ}$ to the field. The force acting on the proton is $(e = 1.6 \times 10^{- 19} C, \sin 60^{\circ} = \cos 30^{\circ} =$ $\frac{\sqrt{3}}{2}$ )

1

1.6 \times 10^{- 12} N

2

3.2 \times 10^{- 12} N

3

4.0 \times 10^{- 12} N

4

2.4 \times 10^{- 12} N

Explanation:

D Given:
$B = 1.732 T$
$v = 10^{7} m / \sec$
$θ = 60^{\circ}$
$e = 1.6 \times 10^{- 19} C$
We know that, force acting on proton
$F_{P} = e (\vec{v} \times \vec{B})$
$= evB \sin θ$
$= evB \sin 60^{\circ}$
$= 1.6 \times 10^{- 19} \times 10^{7} \times 1.732 \times \frac{\sqrt{3}}{2}$
$F_{P} = 2.4 \times 10^{- 12} N$

MHT-CET 2019

Moving Charges & Magnetism

153538 Two particles carrying equal charges move parallel to each other with the speed $150 km / s$ . If $F_{1}$ and $F_{2}$ are magnetic and electric forces between two charged particles then,
$\frac{| F_{1} |}{| F_{2} |} is (Let μ_{0} ε_{0} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2})$

1

1.0 \times 10^{- 6}

2

1.5 \times 10^{- 7}

3

3.0 \times 10^{- 6}

4

2.5 \times 10^{- 7}

Explanation:

D Given, $F_{1} =$ magnetic force $& F_{2} =$ electric force
$v_{1} = v_{2} = 150 km / \sec$
$= 1.5 \times 10^{5} m / \sec$
$μ_{o} ε_{o} = \frac{1}{9 \times 10^{16}} s^{2} / m^{2}$
Electric force between two moving charge particle is
$F_{2} = \frac{μ_{o}}{4 π ε_{o}} \frac{q^{2}}{r^{2}} \dots$
Magnetic force between two moving charge particle is-
$F_{1} = \frac{μ_{o}}{4 π} \frac{q_{1} q_{2} V_{1} V_{2}}{r^{2}}$
$F_{1} = \frac{μ_{o}}{4 π} \frac{q^{2} V^{2}}{r^{2}}$ ...(ii) $(∵ q = q_{2} = q, V_{1} = V_{2} = V)$
Equation (i) is divide by equation (ii)-
$\frac{F_{1}}{F_{2}} = \frac{\frac{μ_{o}}{4 π} \frac{q^{2} v^{2}}{r^{2}}}{\frac{1}{4 π ε_{o}} \frac{q^{2}}{r^{2}}} = ε_{o} μ_{o} v^{2}$
$\frac{F_{1}}{F_{2}} = ε_{0} μ_{0} v^{2}$
$= \frac{1}{9 \times 10^{16}} \times {(1.5 \times 10^{5})}^{2}$
$\frac{F_{1}}{F_{2}} = 2.5 \times 10^{- 7}$

TS- EAMCET-06.05.2019