QBManager

PHXII05:MAGNETISM and MATTER

360396 A bar magnet with a magnetic moment $5.0 A m^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T .$ The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

2 J

3 zero

4

1 J

Explanation:

The work done in turning a magnet from position $θ_{1}$ to $θ_{2}$ in a magnetic field $B$ is,
$W = - Δ U = U_{i} - U_{f}$
$∴ W = M B (\cos θ_{1} - \cos θ_{2})$
As per question, $W = M B (\cos 0^{\circ} - \cos 180^{\circ})$
$W = 2 M B = 2 \times 5.0 \times 0.4 = 4 J$
So, correct option is (1).

JEE - 2023

PHXII05:MAGNETISM and MATTER

360397 A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} J$ of work to turn it through $60^{\circ}$ . The torque needed to maintain the needle in this position will be

1

2 \sqrt{3} J

2

\sqrt{3} J

3

\frac{3}{2} J

4

3 J

Explanation:

$W = m B (\cos θ_{1} - \cos θ_{2})$
$\sqrt{3} = m B (\cos 0^{\circ} - \cos 60^{\circ})$
$m B = 2 \sqrt{3}$
$τ = m B \sin θ = 2 \sqrt{3} \times \sin 60^{\circ} = 3 J$

PHXII05:MAGNETISM and MATTER

360398 A magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field of field strength $H$ . The work done will be

1 2

M H

2

2 π M H

3

M H

4 Zero

Explanation:

Work done to rotate a dipole in field
$\begin{aligned} W = M B (\cos θ_{1} - \cos θ_{2}) \\ θ_{1} = 0^{\circ} and θ_{2} = 360^{\circ} \Rightarrow W = 0 \end{aligned}$

PHXII05:MAGNETISM and MATTER

360399 Assertion :
Magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are parallel to each other.
Reason :
Current loop is treated as a magnetic dipole.

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:

We know that, magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are anti-parallel. A current loop can be treated as a magnetic dipole. When current loop has an area, $A$ and carries a current, $i$ then the magnetic dipole moment is given by $M = i A$ . Option (4) is correct.

PHXII05:MAGNETISM and MATTER

360396 A bar magnet with a magnetic moment $5.0 A m^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T .$ The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

2 J

3 zero

4

1 J

Explanation:

The work done in turning a magnet from position $θ_{1}$ to $θ_{2}$ in a magnetic field $B$ is,
$W = - Δ U = U_{i} - U_{f}$
$∴ W = M B (\cos θ_{1} - \cos θ_{2})$
As per question, $W = M B (\cos 0^{\circ} - \cos 180^{\circ})$
$W = 2 M B = 2 \times 5.0 \times 0.4 = 4 J$
So, correct option is (1).

JEE - 2023

PHXII05:MAGNETISM and MATTER

360397 A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} J$ of work to turn it through $60^{\circ}$ . The torque needed to maintain the needle in this position will be

1

2 \sqrt{3} J

2

\sqrt{3} J

3

\frac{3}{2} J

4

3 J

Explanation:

$W = m B (\cos θ_{1} - \cos θ_{2})$
$\sqrt{3} = m B (\cos 0^{\circ} - \cos 60^{\circ})$
$m B = 2 \sqrt{3}$
$τ = m B \sin θ = 2 \sqrt{3} \times \sin 60^{\circ} = 3 J$

PHXII05:MAGNETISM and MATTER

360398 A magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field of field strength $H$ . The work done will be

1 2

M H

2

2 π M H

3

M H

4 Zero

Explanation:

Work done to rotate a dipole in field
$\begin{aligned} W = M B (\cos θ_{1} - \cos θ_{2}) \\ θ_{1} = 0^{\circ} and θ_{2} = 360^{\circ} \Rightarrow W = 0 \end{aligned}$

PHXII05:MAGNETISM and MATTER

360399 Assertion :
Magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are parallel to each other.
Reason :
Current loop is treated as a magnetic dipole.

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:

We know that, magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are anti-parallel. A current loop can be treated as a magnetic dipole. When current loop has an area, $A$ and carries a current, $i$ then the magnetic dipole moment is given by $M = i A$ . Option (4) is correct.

PHXII05:MAGNETISM and MATTER

360396 A bar magnet with a magnetic moment $5.0 A m^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T .$ The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

2 J

3 zero

4

1 J

Explanation:

The work done in turning a magnet from position $θ_{1}$ to $θ_{2}$ in a magnetic field $B$ is,
$W = - Δ U = U_{i} - U_{f}$
$∴ W = M B (\cos θ_{1} - \cos θ_{2})$
As per question, $W = M B (\cos 0^{\circ} - \cos 180^{\circ})$
$W = 2 M B = 2 \times 5.0 \times 0.4 = 4 J$
So, correct option is (1).

JEE - 2023

PHXII05:MAGNETISM and MATTER

360397 A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} J$ of work to turn it through $60^{\circ}$ . The torque needed to maintain the needle in this position will be

1

2 \sqrt{3} J

2

\sqrt{3} J

3

\frac{3}{2} J

4

3 J

Explanation:

$W = m B (\cos θ_{1} - \cos θ_{2})$
$\sqrt{3} = m B (\cos 0^{\circ} - \cos 60^{\circ})$
$m B = 2 \sqrt{3}$
$τ = m B \sin θ = 2 \sqrt{3} \times \sin 60^{\circ} = 3 J$

PHXII05:MAGNETISM and MATTER

360398 A magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field of field strength $H$ . The work done will be

1 2

M H

2

2 π M H

3

M H

4 Zero

Explanation:

Work done to rotate a dipole in field
$\begin{aligned} W = M B (\cos θ_{1} - \cos θ_{2}) \\ θ_{1} = 0^{\circ} and θ_{2} = 360^{\circ} \Rightarrow W = 0 \end{aligned}$

PHXII05:MAGNETISM and MATTER

360399 Assertion :
Magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are parallel to each other.
Reason :
Current loop is treated as a magnetic dipole.

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:

We know that, magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are anti-parallel. A current loop can be treated as a magnetic dipole. When current loop has an area, $A$ and carries a current, $i$ then the magnetic dipole moment is given by $M = i A$ . Option (4) is correct.

PHXII05:MAGNETISM and MATTER

360396 A bar magnet with a magnetic moment $5.0 A m^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T .$ The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

1

4 J

2

2 J

3 zero

4

1 J

Explanation:

The work done in turning a magnet from position $θ_{1}$ to $θ_{2}$ in a magnetic field $B$ is,
$W = - Δ U = U_{i} - U_{f}$
$∴ W = M B (\cos θ_{1} - \cos θ_{2})$
As per question, $W = M B (\cos 0^{\circ} - \cos 180^{\circ})$
$W = 2 M B = 2 \times 5.0 \times 0.4 = 4 J$
So, correct option is (1).

JEE - 2023

PHXII05:MAGNETISM and MATTER

360397 A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} J$ of work to turn it through $60^{\circ}$ . The torque needed to maintain the needle in this position will be

1

2 \sqrt{3} J

2

\sqrt{3} J

3

\frac{3}{2} J

4

3 J

Explanation:

$W = m B (\cos θ_{1} - \cos θ_{2})$
$\sqrt{3} = m B (\cos 0^{\circ} - \cos 60^{\circ})$
$m B = 2 \sqrt{3}$
$τ = m B \sin θ = 2 \sqrt{3} \times \sin 60^{\circ} = 3 J$

PHXII05:MAGNETISM and MATTER

360398 A magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field of field strength $H$ . The work done will be

1 2

M H

2

2 π M H

3

M H

4 Zero

Explanation:

Work done to rotate a dipole in field
$\begin{aligned} W = M B (\cos θ_{1} - \cos θ_{2}) \\ θ_{1} = 0^{\circ} and θ_{2} = 360^{\circ} \Rightarrow W = 0 \end{aligned}$

PHXII05:MAGNETISM and MATTER

360399 Assertion :
Magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are parallel to each other.
Reason :
Current loop is treated as a magnetic dipole.

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:

We know that, magnetic dipole possesses maximum potential energy when magnetic moment and magnetic field are anti-parallel. A current loop can be treated as a magnetic dipole. When current loop has an area, $A$ and carries a current, $i$ then the magnetic dipole moment is given by $M = i A$ . Option (4) is correct.

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: