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PHXII04:MOVING CHARGES AND MAGNETISM

362746 Two long current carrying thin wires, both with current $I$ , are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $θ$ with the vertical. If wires have mass $λ$ per unit length then the value of $I$ is: ( $g =$ gravitational acceleration)
supporting img

1

2 \sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

2

2 \sqrt{\frac{π g L}{μ_{0}} \tan θ}

3

\sqrt{\frac{π λ g L}{μ_{0}} \tan θ}

4

\sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

Explanation:

Let $T$ is the tension in the strings and $F_{B}$ is the magnetic force on the wire of length $L$ .
supporting img
At equilibrium
$\begin{aligned} T \cos θ = F_{g} = λ L g \\ T \sin θ = I L \frac{μ_{o} I}{2 π d} \end{aligned}$
From the above equations we get
$I = 2 \sin θ \sqrt{\frac{π λ g L}{μ_{o} \cos θ}} (d = 2 L \sin θ)$

PHXII04:MOVING CHARGES AND MAGNETISM

362747 A metallic rod of mass per unit length $0.5 k g m^{- 1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 $T$ acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

1 14.76

A

2 11.32

A

3 5.98

A

4 7.14

A

Explanation:

supporting img
$F = i (\vec{L} \times \vec{B}) = i L B \sin 90^{\circ}$
$F \cos 30^{\circ} = m g \sin 30^{\circ}$
$B i L \cos 30^{\circ} = m g \sin 30^{\circ}$
$i = \frac{m g \tan 30^{\circ}}{B L} = \frac{0.5 \times 9.8 \times \frac{1}{\sqrt{3}}}{0.25} = \frac{19.6}{1.73} = 11.3 A$

NEET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362748 A uniform magnetic field of 1.5 $T$ exists in cylindrical region of radius 10 $c m$ . Its direction parallel to the axis along east to west. A wire carrying current of 7 $A$ in the north east to south west direction passes through this rigion. The magnitude and direction of the force on the wire is

1 2.1

N

vertically upwards

2 2.1

N

vertically downwards

3 4.1

N

vertically upwards

4 4.1

N

vertically downwards

Explanation:

Angle between magnetic field and current is $π / 4$ .
supporting img

length $AB$ of the wire inside the magnetic field,
$l = \frac{2 r}{\sin 45^{\circ}} = \frac{2 \times 0.1}{\sin 45^{\circ}}$
$\begin{aligned} F = Bil \sin θ & = 1.5 \times 7 \times \frac{2 \times 0.1}{\sin 45^{\circ}} \times \sin 45^{\circ} \\ = 2.1 N (vertically downwards) \end{aligned}$

NCERT Exemplar

PHXII04:MOVING CHARGES AND MAGNETISM

362749 Given below two statements One is labelled as ( $A$ ) and the other is labelled as Reason( $R$ ).
Assertion :
In an uniform magnetic field, speed and energy remains the same for a moving charged particle.
Reason :
Moving charged particle experiences magnetic force perpendicular to its direction of motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v$ remains same if q,m, $\vec{B}$ remain same
$\vec{F} = q (\vec{v} \times \vec{B})$
As it is cross product of vectors,
$\vec{F} ⊥ \vec{v} \Rightarrow θ = 90^{\circ}$
Here, Work done $= \vec{F} \cdot \vec{S}$ (Note that this is dot product)
$= FS \cos 90^{\circ} = 0$ .
Work done $= 0$

JEE - 2022

PHXII04:MOVING CHARGES AND MAGNETISM

362746 Two long current carrying thin wires, both with current $I$ , are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $θ$ with the vertical. If wires have mass $λ$ per unit length then the value of $I$ is: ( $g =$ gravitational acceleration)
supporting img

1

2 \sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

2

2 \sqrt{\frac{π g L}{μ_{0}} \tan θ}

3

\sqrt{\frac{π λ g L}{μ_{0}} \tan θ}

4

\sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

Explanation:

Let $T$ is the tension in the strings and $F_{B}$ is the magnetic force on the wire of length $L$ .
supporting img
At equilibrium
$\begin{aligned} T \cos θ = F_{g} = λ L g \\ T \sin θ = I L \frac{μ_{o} I}{2 π d} \end{aligned}$
From the above equations we get
$I = 2 \sin θ \sqrt{\frac{π λ g L}{μ_{o} \cos θ}} (d = 2 L \sin θ)$

PHXII04:MOVING CHARGES AND MAGNETISM

362747 A metallic rod of mass per unit length $0.5 k g m^{- 1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 $T$ acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

1 14.76

A

2 11.32

A

3 5.98

A

4 7.14

A

Explanation:

supporting img
$F = i (\vec{L} \times \vec{B}) = i L B \sin 90^{\circ}$
$F \cos 30^{\circ} = m g \sin 30^{\circ}$
$B i L \cos 30^{\circ} = m g \sin 30^{\circ}$
$i = \frac{m g \tan 30^{\circ}}{B L} = \frac{0.5 \times 9.8 \times \frac{1}{\sqrt{3}}}{0.25} = \frac{19.6}{1.73} = 11.3 A$

NEET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362748 A uniform magnetic field of 1.5 $T$ exists in cylindrical region of radius 10 $c m$ . Its direction parallel to the axis along east to west. A wire carrying current of 7 $A$ in the north east to south west direction passes through this rigion. The magnitude and direction of the force on the wire is

1 2.1

N

vertically upwards

2 2.1

N

vertically downwards

3 4.1

N

vertically upwards

4 4.1

N

vertically downwards

Explanation:

Angle between magnetic field and current is $π / 4$ .
supporting img

length $AB$ of the wire inside the magnetic field,
$l = \frac{2 r}{\sin 45^{\circ}} = \frac{2 \times 0.1}{\sin 45^{\circ}}$
$\begin{aligned} F = Bil \sin θ & = 1.5 \times 7 \times \frac{2 \times 0.1}{\sin 45^{\circ}} \times \sin 45^{\circ} \\ = 2.1 N (vertically downwards) \end{aligned}$

NCERT Exemplar

PHXII04:MOVING CHARGES AND MAGNETISM

362749 Given below two statements One is labelled as ( $A$ ) and the other is labelled as Reason( $R$ ).
Assertion :
In an uniform magnetic field, speed and energy remains the same for a moving charged particle.
Reason :
Moving charged particle experiences magnetic force perpendicular to its direction of motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v$ remains same if q,m, $\vec{B}$ remain same
$\vec{F} = q (\vec{v} \times \vec{B})$
As it is cross product of vectors,
$\vec{F} ⊥ \vec{v} \Rightarrow θ = 90^{\circ}$
Here, Work done $= \vec{F} \cdot \vec{S}$ (Note that this is dot product)
$= FS \cos 90^{\circ} = 0$ .
Work done $= 0$

JEE - 2022

PHXII04:MOVING CHARGES AND MAGNETISM

362746 Two long current carrying thin wires, both with current $I$ , are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $θ$ with the vertical. If wires have mass $λ$ per unit length then the value of $I$ is: ( $g =$ gravitational acceleration)
supporting img

1

2 \sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

2

2 \sqrt{\frac{π g L}{μ_{0}} \tan θ}

3

\sqrt{\frac{π λ g L}{μ_{0}} \tan θ}

4

\sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

Explanation:

Let $T$ is the tension in the strings and $F_{B}$ is the magnetic force on the wire of length $L$ .
supporting img
At equilibrium
$\begin{aligned} T \cos θ = F_{g} = λ L g \\ T \sin θ = I L \frac{μ_{o} I}{2 π d} \end{aligned}$
From the above equations we get
$I = 2 \sin θ \sqrt{\frac{π λ g L}{μ_{o} \cos θ}} (d = 2 L \sin θ)$

PHXII04:MOVING CHARGES AND MAGNETISM

362747 A metallic rod of mass per unit length $0.5 k g m^{- 1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 $T$ acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

1 14.76

A

2 11.32

A

3 5.98

A

4 7.14

A

Explanation:

supporting img
$F = i (\vec{L} \times \vec{B}) = i L B \sin 90^{\circ}$
$F \cos 30^{\circ} = m g \sin 30^{\circ}$
$B i L \cos 30^{\circ} = m g \sin 30^{\circ}$
$i = \frac{m g \tan 30^{\circ}}{B L} = \frac{0.5 \times 9.8 \times \frac{1}{\sqrt{3}}}{0.25} = \frac{19.6}{1.73} = 11.3 A$

NEET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362748 A uniform magnetic field of 1.5 $T$ exists in cylindrical region of radius 10 $c m$ . Its direction parallel to the axis along east to west. A wire carrying current of 7 $A$ in the north east to south west direction passes through this rigion. The magnitude and direction of the force on the wire is

1 2.1

N

vertically upwards

2 2.1

N

vertically downwards

3 4.1

N

vertically upwards

4 4.1

N

vertically downwards

Explanation:

Angle between magnetic field and current is $π / 4$ .
supporting img

length $AB$ of the wire inside the magnetic field,
$l = \frac{2 r}{\sin 45^{\circ}} = \frac{2 \times 0.1}{\sin 45^{\circ}}$
$\begin{aligned} F = Bil \sin θ & = 1.5 \times 7 \times \frac{2 \times 0.1}{\sin 45^{\circ}} \times \sin 45^{\circ} \\ = 2.1 N (vertically downwards) \end{aligned}$

NCERT Exemplar

PHXII04:MOVING CHARGES AND MAGNETISM

362749 Given below two statements One is labelled as ( $A$ ) and the other is labelled as Reason( $R$ ).
Assertion :
In an uniform magnetic field, speed and energy remains the same for a moving charged particle.
Reason :
Moving charged particle experiences magnetic force perpendicular to its direction of motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v$ remains same if q,m, $\vec{B}$ remain same
$\vec{F} = q (\vec{v} \times \vec{B})$
As it is cross product of vectors,
$\vec{F} ⊥ \vec{v} \Rightarrow θ = 90^{\circ}$
Here, Work done $= \vec{F} \cdot \vec{S}$ (Note that this is dot product)
$= FS \cos 90^{\circ} = 0$ .
Work done $= 0$

JEE - 2022

PHXII04:MOVING CHARGES AND MAGNETISM

362746 Two long current carrying thin wires, both with current $I$ , are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $θ$ with the vertical. If wires have mass $λ$ per unit length then the value of $I$ is: ( $g =$ gravitational acceleration)
supporting img

1

2 \sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

2

2 \sqrt{\frac{π g L}{μ_{0}} \tan θ}

3

\sqrt{\frac{π λ g L}{μ_{0}} \tan θ}

4

\sin θ \sqrt{\frac{π λ g L}{μ_{0} \cos θ}}

Explanation:

Let $T$ is the tension in the strings and $F_{B}$ is the magnetic force on the wire of length $L$ .
supporting img
At equilibrium
$\begin{aligned} T \cos θ = F_{g} = λ L g \\ T \sin θ = I L \frac{μ_{o} I}{2 π d} \end{aligned}$
From the above equations we get
$I = 2 \sin θ \sqrt{\frac{π λ g L}{μ_{o} \cos θ}} (d = 2 L \sin θ)$

PHXII04:MOVING CHARGES AND MAGNETISM

362747 A metallic rod of mass per unit length $0.5 k g m^{- 1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 $T$ acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

1 14.76

A

2 11.32

A

3 5.98

A

4 7.14

A

Explanation:

supporting img
$F = i (\vec{L} \times \vec{B}) = i L B \sin 90^{\circ}$
$F \cos 30^{\circ} = m g \sin 30^{\circ}$
$B i L \cos 30^{\circ} = m g \sin 30^{\circ}$
$i = \frac{m g \tan 30^{\circ}}{B L} = \frac{0.5 \times 9.8 \times \frac{1}{\sqrt{3}}}{0.25} = \frac{19.6}{1.73} = 11.3 A$

NEET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362748 A uniform magnetic field of 1.5 $T$ exists in cylindrical region of radius 10 $c m$ . Its direction parallel to the axis along east to west. A wire carrying current of 7 $A$ in the north east to south west direction passes through this rigion. The magnitude and direction of the force on the wire is

1 2.1

N

vertically upwards

2 2.1

N

vertically downwards

3 4.1

N

vertically upwards

4 4.1

N

vertically downwards

Explanation:

Angle between magnetic field and current is $π / 4$ .
supporting img

length $AB$ of the wire inside the magnetic field,
$l = \frac{2 r}{\sin 45^{\circ}} = \frac{2 \times 0.1}{\sin 45^{\circ}}$
$\begin{aligned} F = Bil \sin θ & = 1.5 \times 7 \times \frac{2 \times 0.1}{\sin 45^{\circ}} \times \sin 45^{\circ} \\ = 2.1 N (vertically downwards) \end{aligned}$

NCERT Exemplar

PHXII04:MOVING CHARGES AND MAGNETISM

362749 Given below two statements One is labelled as ( $A$ ) and the other is labelled as Reason( $R$ ).
Assertion :
In an uniform magnetic field, speed and energy remains the same for a moving charged particle.
Reason :
Moving charged particle experiences magnetic force perpendicular to its direction of motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v$ remains same if q,m, $\vec{B}$ remain same
$\vec{F} = q (\vec{v} \times \vec{B})$
As it is cross product of vectors,
$\vec{F} ⊥ \vec{v} \Rightarrow θ = 90^{\circ}$
Here, Work done $= \vec{F} \cdot \vec{S}$ (Note that this is dot product)
$= FS \cos 90^{\circ} = 0$ .
Work done $= 0$

JEE - 2022