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Moving Charges & Magnetism

153690 A massless square loop, of wire of resistance $10 Ω$ , supporting a mass of $1 g$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , directed outwards in the shaded region. A de voltage $V$ is applied to the loop. For what value of $V$ , the magnetic force will exactly balance the weight of the supporting mass of $1 g$ ? (If sides of the $100 p = 10 cm, g = 10$ ${ms}^{- 2}$ )

1

\frac{1}{10} V

2

1 V

3

100 V

4

10 V

Explanation:

D Given,
Magnetic field $(B) = 10^{3}$ Gauss $= 10^{3} \times 10^{- 4} T = 10^{- 1} T$
Resistance of wire $(R) = 10 Ω$
Side of the loop $(l) = 10 cm = 10 \times 10^{- 2} m$
Mass $(m) = 1 gm = 1 \times 10^{- 3} kg$
We know that,
$F = i / B$
$mg = i / B$
$1 \times 10^{- 3} \times 10 = \frac{V}{R} \times 10 \times 10^{- 2} \times 10^{- 1}$
$1 = \frac{V}{10}$
So, $V = 10$ volt

JEE Main-30.01.2023

Moving Charges & Magnetism

153692 A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with $B = 12$ $m T$ . The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is:
(Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

192 k {Vm}^{- 1}

2

192 m {Vm}^{- 1}

3

9600 k {Vm}^{- 1}

4

16 k {Vm}^{- 1}

Explanation:

A Given,
$\vec{E} = E \hat{k}$
$\vec{B} = Bj$ and $B = 12 mT = 12 \times 10^{- 3} T$
Energy $(E) = 728 eV = 728 \times 1.6 \times 10^{- 19} V$
We know that, $E = \frac{1}{2} {mv}^{2}$
$728 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2}$
$v = 16 \times 10^{6} m / \sec$
$∴ E = v \times B = 16 \times 10^{6} \times 12 \times 10^{- 3}$
$E = 192 \times 10^{3} V / m$

JEE Main-26.07.2022

Moving Charges & Magnetism

153694 Statement (I): A uniform electric field and a uniform magnetic field are pointed in the same direction. If an electron is projected in the same direction the electron velocity will decrease in magnitude.
Statement (II): Two infinite long parallel wires are carrying current in the same direction. The magnetic field at a point midway between the wires is zero.
Statement (III): No net force acts on a rectangular coil carrying a steady current when suspended in a uniform magnetic field.
Which of the following is correct?

1 Statement I, II and III are true.

2 Statement I and II are true, but statement III is false.

3 Statement II and III are true, but statement I is false.

4 Statement I, III are true, but Statement II is false.

TS EAMCET 19.07.2022

Moving Charges & Magnetism

153696 Given the fact that
A) Magnetic field exerts force only on a moving charge
B) Electric field exerts force on both stationary and moving charge C) Magnetic field exerts force on charge moving parallel to the direction of the field.
Which of the following is true?

1 A and

C

2 A and B

3

B

and

C

4 A, B and C

Explanation:

B As we know that, magnetic field (B) produces only due to current i.e. flow of charge per unit time.
But electric field is generated by any charged particle no matter it is static or moving.
Hence, we can say magnetic field exerts force only on moving charge but electric field exerts force on moving as well as static both.
Force on a charged particle, $F = qvB \sin θ$
When motion is parallel to the direction of magnetic field,
Then, $θ = 0^{\circ}$ or $180^{\circ}$
$\sin 0^{\circ} = 0 and \sin 180^{\circ} = 0$
So, $F = 0$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153697 The electron is hydrogen atom is moving in an orbit of radius $0.53 \AA$ . It takes $1.571 \times 10^{- 16} s$ to complete one revolution. The velocity of electron will be $π = 3.142$

1

3 \times 10^{8} \frac{m}{s}

2

4 \times 10^{6} \frac{m}{s}

3

2.12 \times 10^{6} \frac{m}{s}

4

5.3 \times 10^{6} \frac{m}{s}

Explanation:

C Given,
Radius of orbit $= 0.53 \AA = 0.53 \times 10^{- 10} m$
Time $(t) = 1.571 \times 10^{- 16} s$
We know that, $v = \frac{2 π r}{t}$
$v = \frac{2 \times 3.14 \times 0.53 \times 10^{- 10}}{1.571 \times 10^{- 16}}$
$v = 2.12 \times 10^{6} m / s$

MHT-CET 2020

Moving Charges & Magnetism

153690 A massless square loop, of wire of resistance $10 Ω$ , supporting a mass of $1 g$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , directed outwards in the shaded region. A de voltage $V$ is applied to the loop. For what value of $V$ , the magnetic force will exactly balance the weight of the supporting mass of $1 g$ ? (If sides of the $100 p = 10 cm, g = 10$ ${ms}^{- 2}$ )

1

\frac{1}{10} V

2

1 V

3

100 V

4

10 V

Explanation:

D Given,
Magnetic field $(B) = 10^{3}$ Gauss $= 10^{3} \times 10^{- 4} T = 10^{- 1} T$
Resistance of wire $(R) = 10 Ω$
Side of the loop $(l) = 10 cm = 10 \times 10^{- 2} m$
Mass $(m) = 1 gm = 1 \times 10^{- 3} kg$
We know that,
$F = i / B$
$mg = i / B$
$1 \times 10^{- 3} \times 10 = \frac{V}{R} \times 10 \times 10^{- 2} \times 10^{- 1}$
$1 = \frac{V}{10}$
So, $V = 10$ volt

JEE Main-30.01.2023

Moving Charges & Magnetism

153692 A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with $B = 12$ $m T$ . The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is:
(Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

192 k {Vm}^{- 1}

2

192 m {Vm}^{- 1}

3

9600 k {Vm}^{- 1}

4

16 k {Vm}^{- 1}

Explanation:

A Given,
$\vec{E} = E \hat{k}$
$\vec{B} = Bj$ and $B = 12 mT = 12 \times 10^{- 3} T$
Energy $(E) = 728 eV = 728 \times 1.6 \times 10^{- 19} V$
We know that, $E = \frac{1}{2} {mv}^{2}$
$728 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2}$
$v = 16 \times 10^{6} m / \sec$
$∴ E = v \times B = 16 \times 10^{6} \times 12 \times 10^{- 3}$
$E = 192 \times 10^{3} V / m$

JEE Main-26.07.2022

Moving Charges & Magnetism

153694 Statement (I): A uniform electric field and a uniform magnetic field are pointed in the same direction. If an electron is projected in the same direction the electron velocity will decrease in magnitude.
Statement (II): Two infinite long parallel wires are carrying current in the same direction. The magnetic field at a point midway between the wires is zero.
Statement (III): No net force acts on a rectangular coil carrying a steady current when suspended in a uniform magnetic field.
Which of the following is correct?

1 Statement I, II and III are true.

2 Statement I and II are true, but statement III is false.

3 Statement II and III are true, but statement I is false.

4 Statement I, III are true, but Statement II is false.

TS EAMCET 19.07.2022

Moving Charges & Magnetism

153696 Given the fact that
A) Magnetic field exerts force only on a moving charge
B) Electric field exerts force on both stationary and moving charge C) Magnetic field exerts force on charge moving parallel to the direction of the field.
Which of the following is true?

1 A and

C

2 A and B

3

B

and

C

4 A, B and C

Explanation:

B As we know that, magnetic field (B) produces only due to current i.e. flow of charge per unit time.
But electric field is generated by any charged particle no matter it is static or moving.
Hence, we can say magnetic field exerts force only on moving charge but electric field exerts force on moving as well as static both.
Force on a charged particle, $F = qvB \sin θ$
When motion is parallel to the direction of magnetic field,
Then, $θ = 0^{\circ}$ or $180^{\circ}$
$\sin 0^{\circ} = 0 and \sin 180^{\circ} = 0$
So, $F = 0$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153697 The electron is hydrogen atom is moving in an orbit of radius $0.53 \AA$ . It takes $1.571 \times 10^{- 16} s$ to complete one revolution. The velocity of electron will be $π = 3.142$

1

3 \times 10^{8} \frac{m}{s}

2

4 \times 10^{6} \frac{m}{s}

3

2.12 \times 10^{6} \frac{m}{s}

4

5.3 \times 10^{6} \frac{m}{s}

Explanation:

C Given,
Radius of orbit $= 0.53 \AA = 0.53 \times 10^{- 10} m$
Time $(t) = 1.571 \times 10^{- 16} s$
We know that, $v = \frac{2 π r}{t}$
$v = \frac{2 \times 3.14 \times 0.53 \times 10^{- 10}}{1.571 \times 10^{- 16}}$
$v = 2.12 \times 10^{6} m / s$

MHT-CET 2020

Moving Charges & Magnetism

153690 A massless square loop, of wire of resistance $10 Ω$ , supporting a mass of $1 g$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , directed outwards in the shaded region. A de voltage $V$ is applied to the loop. For what value of $V$ , the magnetic force will exactly balance the weight of the supporting mass of $1 g$ ? (If sides of the $100 p = 10 cm, g = 10$ ${ms}^{- 2}$ )

1

\frac{1}{10} V

2

1 V

3

100 V

4

10 V

Explanation:

D Given,
Magnetic field $(B) = 10^{3}$ Gauss $= 10^{3} \times 10^{- 4} T = 10^{- 1} T$
Resistance of wire $(R) = 10 Ω$
Side of the loop $(l) = 10 cm = 10 \times 10^{- 2} m$
Mass $(m) = 1 gm = 1 \times 10^{- 3} kg$
We know that,
$F = i / B$
$mg = i / B$
$1 \times 10^{- 3} \times 10 = \frac{V}{R} \times 10 \times 10^{- 2} \times 10^{- 1}$
$1 = \frac{V}{10}$
So, $V = 10$ volt

JEE Main-30.01.2023

Moving Charges & Magnetism

153692 A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with $B = 12$ $m T$ . The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is:
(Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

192 k {Vm}^{- 1}

2

192 m {Vm}^{- 1}

3

9600 k {Vm}^{- 1}

4

16 k {Vm}^{- 1}

Explanation:

A Given,
$\vec{E} = E \hat{k}$
$\vec{B} = Bj$ and $B = 12 mT = 12 \times 10^{- 3} T$
Energy $(E) = 728 eV = 728 \times 1.6 \times 10^{- 19} V$
We know that, $E = \frac{1}{2} {mv}^{2}$
$728 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2}$
$v = 16 \times 10^{6} m / \sec$
$∴ E = v \times B = 16 \times 10^{6} \times 12 \times 10^{- 3}$
$E = 192 \times 10^{3} V / m$

JEE Main-26.07.2022

Moving Charges & Magnetism

153694 Statement (I): A uniform electric field and a uniform magnetic field are pointed in the same direction. If an electron is projected in the same direction the electron velocity will decrease in magnitude.
Statement (II): Two infinite long parallel wires are carrying current in the same direction. The magnetic field at a point midway between the wires is zero.
Statement (III): No net force acts on a rectangular coil carrying a steady current when suspended in a uniform magnetic field.
Which of the following is correct?

1 Statement I, II and III are true.

2 Statement I and II are true, but statement III is false.

3 Statement II and III are true, but statement I is false.

4 Statement I, III are true, but Statement II is false.

TS EAMCET 19.07.2022

Moving Charges & Magnetism

153696 Given the fact that
A) Magnetic field exerts force only on a moving charge
B) Electric field exerts force on both stationary and moving charge C) Magnetic field exerts force on charge moving parallel to the direction of the field.
Which of the following is true?

1 A and

C

2 A and B

3

B

and

C

4 A, B and C

Explanation:

B As we know that, magnetic field (B) produces only due to current i.e. flow of charge per unit time.
But electric field is generated by any charged particle no matter it is static or moving.
Hence, we can say magnetic field exerts force only on moving charge but electric field exerts force on moving as well as static both.
Force on a charged particle, $F = qvB \sin θ$
When motion is parallel to the direction of magnetic field,
Then, $θ = 0^{\circ}$ or $180^{\circ}$
$\sin 0^{\circ} = 0 and \sin 180^{\circ} = 0$
So, $F = 0$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153697 The electron is hydrogen atom is moving in an orbit of radius $0.53 \AA$ . It takes $1.571 \times 10^{- 16} s$ to complete one revolution. The velocity of electron will be $π = 3.142$

1

3 \times 10^{8} \frac{m}{s}

2

4 \times 10^{6} \frac{m}{s}

3

2.12 \times 10^{6} \frac{m}{s}

4

5.3 \times 10^{6} \frac{m}{s}

Explanation:

C Given,
Radius of orbit $= 0.53 \AA = 0.53 \times 10^{- 10} m$
Time $(t) = 1.571 \times 10^{- 16} s$
We know that, $v = \frac{2 π r}{t}$
$v = \frac{2 \times 3.14 \times 0.53 \times 10^{- 10}}{1.571 \times 10^{- 16}}$
$v = 2.12 \times 10^{6} m / s$

MHT-CET 2020

Moving Charges & Magnetism

153690 A massless square loop, of wire of resistance $10 Ω$ , supporting a mass of $1 g$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , directed outwards in the shaded region. A de voltage $V$ is applied to the loop. For what value of $V$ , the magnetic force will exactly balance the weight of the supporting mass of $1 g$ ? (If sides of the $100 p = 10 cm, g = 10$ ${ms}^{- 2}$ )

1

\frac{1}{10} V

2

1 V

3

100 V

4

10 V

Explanation:

D Given,
Magnetic field $(B) = 10^{3}$ Gauss $= 10^{3} \times 10^{- 4} T = 10^{- 1} T$
Resistance of wire $(R) = 10 Ω$
Side of the loop $(l) = 10 cm = 10 \times 10^{- 2} m$
Mass $(m) = 1 gm = 1 \times 10^{- 3} kg$
We know that,
$F = i / B$
$mg = i / B$
$1 \times 10^{- 3} \times 10 = \frac{V}{R} \times 10 \times 10^{- 2} \times 10^{- 1}$
$1 = \frac{V}{10}$
So, $V = 10$ volt

JEE Main-30.01.2023

Moving Charges & Magnetism

153692 A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with $B = 12$ $m T$ . The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is:
(Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

192 k {Vm}^{- 1}

2

192 m {Vm}^{- 1}

3

9600 k {Vm}^{- 1}

4

16 k {Vm}^{- 1}

Explanation:

A Given,
$\vec{E} = E \hat{k}$
$\vec{B} = Bj$ and $B = 12 mT = 12 \times 10^{- 3} T$
Energy $(E) = 728 eV = 728 \times 1.6 \times 10^{- 19} V$
We know that, $E = \frac{1}{2} {mv}^{2}$
$728 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2}$
$v = 16 \times 10^{6} m / \sec$
$∴ E = v \times B = 16 \times 10^{6} \times 12 \times 10^{- 3}$
$E = 192 \times 10^{3} V / m$

JEE Main-26.07.2022

Moving Charges & Magnetism

153694 Statement (I): A uniform electric field and a uniform magnetic field are pointed in the same direction. If an electron is projected in the same direction the electron velocity will decrease in magnitude.
Statement (II): Two infinite long parallel wires are carrying current in the same direction. The magnetic field at a point midway between the wires is zero.
Statement (III): No net force acts on a rectangular coil carrying a steady current when suspended in a uniform magnetic field.
Which of the following is correct?

1 Statement I, II and III are true.

2 Statement I and II are true, but statement III is false.

3 Statement II and III are true, but statement I is false.

4 Statement I, III are true, but Statement II is false.

TS EAMCET 19.07.2022

Moving Charges & Magnetism

153696 Given the fact that
A) Magnetic field exerts force only on a moving charge
B) Electric field exerts force on both stationary and moving charge C) Magnetic field exerts force on charge moving parallel to the direction of the field.
Which of the following is true?

1 A and

C

2 A and B

3

B

and

C

4 A, B and C

Explanation:

B As we know that, magnetic field (B) produces only due to current i.e. flow of charge per unit time.
But electric field is generated by any charged particle no matter it is static or moving.
Hence, we can say magnetic field exerts force only on moving charge but electric field exerts force on moving as well as static both.
Force on a charged particle, $F = qvB \sin θ$
When motion is parallel to the direction of magnetic field,
Then, $θ = 0^{\circ}$ or $180^{\circ}$
$\sin 0^{\circ} = 0 and \sin 180^{\circ} = 0$
So, $F = 0$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153697 The electron is hydrogen atom is moving in an orbit of radius $0.53 \AA$ . It takes $1.571 \times 10^{- 16} s$ to complete one revolution. The velocity of electron will be $π = 3.142$

1

3 \times 10^{8} \frac{m}{s}

2

4 \times 10^{6} \frac{m}{s}

3

2.12 \times 10^{6} \frac{m}{s}

4

5.3 \times 10^{6} \frac{m}{s}

Explanation:

C Given,
Radius of orbit $= 0.53 \AA = 0.53 \times 10^{- 10} m$
Time $(t) = 1.571 \times 10^{- 16} s$
We know that, $v = \frac{2 π r}{t}$
$v = \frac{2 \times 3.14 \times 0.53 \times 10^{- 10}}{1.571 \times 10^{- 16}}$
$v = 2.12 \times 10^{6} m / s$

MHT-CET 2020

Moving Charges & Magnetism

153690 A massless square loop, of wire of resistance $10 Ω$ , supporting a mass of $1 g$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , directed outwards in the shaded region. A de voltage $V$ is applied to the loop. For what value of $V$ , the magnetic force will exactly balance the weight of the supporting mass of $1 g$ ? (If sides of the $100 p = 10 cm, g = 10$ ${ms}^{- 2}$ )

1

\frac{1}{10} V

2

1 V

3

100 V

4

10 V

Explanation:

D Given,
Magnetic field $(B) = 10^{3}$ Gauss $= 10^{3} \times 10^{- 4} T = 10^{- 1} T$
Resistance of wire $(R) = 10 Ω$
Side of the loop $(l) = 10 cm = 10 \times 10^{- 2} m$
Mass $(m) = 1 gm = 1 \times 10^{- 3} kg$
We know that,
$F = i / B$
$mg = i / B$
$1 \times 10^{- 3} \times 10 = \frac{V}{R} \times 10 \times 10^{- 2} \times 10^{- 1}$
$1 = \frac{V}{10}$
So, $V = 10$ volt

JEE Main-30.01.2023

Moving Charges & Magnetism

153692 A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with $B = 12$ $m T$ . The value $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is:
(Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

192 k {Vm}^{- 1}

2

192 m {Vm}^{- 1}

3

9600 k {Vm}^{- 1}

4

16 k {Vm}^{- 1}

Explanation:

A Given,
$\vec{E} = E \hat{k}$
$\vec{B} = Bj$ and $B = 12 mT = 12 \times 10^{- 3} T$
Energy $(E) = 728 eV = 728 \times 1.6 \times 10^{- 19} V$
We know that, $E = \frac{1}{2} {mv}^{2}$
$728 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2}$
$v = 16 \times 10^{6} m / \sec$
$∴ E = v \times B = 16 \times 10^{6} \times 12 \times 10^{- 3}$
$E = 192 \times 10^{3} V / m$

JEE Main-26.07.2022

Moving Charges & Magnetism

153694 Statement (I): A uniform electric field and a uniform magnetic field are pointed in the same direction. If an electron is projected in the same direction the electron velocity will decrease in magnitude.
Statement (II): Two infinite long parallel wires are carrying current in the same direction. The magnetic field at a point midway between the wires is zero.
Statement (III): No net force acts on a rectangular coil carrying a steady current when suspended in a uniform magnetic field.
Which of the following is correct?

1 Statement I, II and III are true.

2 Statement I and II are true, but statement III is false.

3 Statement II and III are true, but statement I is false.

4 Statement I, III are true, but Statement II is false.

TS EAMCET 19.07.2022

Moving Charges & Magnetism

153696 Given the fact that
A) Magnetic field exerts force only on a moving charge
B) Electric field exerts force on both stationary and moving charge C) Magnetic field exerts force on charge moving parallel to the direction of the field.
Which of the following is true?

1 A and

C

2 A and B

3

B

and

C

4 A, B and C

Explanation:

B As we know that, magnetic field (B) produces only due to current i.e. flow of charge per unit time.
But electric field is generated by any charged particle no matter it is static or moving.
Hence, we can say magnetic field exerts force only on moving charge but electric field exerts force on moving as well as static both.
Force on a charged particle, $F = qvB \sin θ$
When motion is parallel to the direction of magnetic field,
Then, $θ = 0^{\circ}$ or $180^{\circ}$
$\sin 0^{\circ} = 0 and \sin 180^{\circ} = 0$
So, $F = 0$

TS EAMCET 05.08.2021

Moving Charges & Magnetism

153697 The electron is hydrogen atom is moving in an orbit of radius $0.53 \AA$ . It takes $1.571 \times 10^{- 16} s$ to complete one revolution. The velocity of electron will be $π = 3.142$

1

3 \times 10^{8} \frac{m}{s}

2

4 \times 10^{6} \frac{m}{s}

3

2.12 \times 10^{6} \frac{m}{s}

4

5.3 \times 10^{6} \frac{m}{s}

Explanation:

C Given,
Radius of orbit $= 0.53 \AA = 0.53 \times 10^{- 10} m$
Time $(t) = 1.571 \times 10^{- 16} s$
We know that, $v = \frac{2 π r}{t}$
$v = \frac{2 \times 3.14 \times 0.53 \times 10^{- 10}}{1.571 \times 10^{- 16}}$
$v = 2.12 \times 10^{6} m / s$

MHT-CET 2020

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