A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is(are)

In the region 0 2R, the magnetic field is non-zero.

In the region 0 < r < R, the magnetic field is non-zero.

In the region R < r < 2R, the magnetic field is along the common axis.

In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centred on the axis.

In the region r > 2R, the magnetic field is non-zero.

Magnetism PYQ — JEE Advanced 2013 · Shift 2 · Q48 with Solution

View written solutionFree

Correct answer: A, D

The total magnetic field at any point P is the vector sum of the magnetic field due to the hollow cylindrical conductor ( $B_{cyl}$ ) and the magnetic field due to the solenoid ( $B_{sol}$ ). We will analyze the magnetic field in each region specified by the options.

1. Magnetic Field of the Infinite Solenoid ( $B_{sol}$ )

For an ideal infinite solenoid with radius $R_{sol} = 2R$ and $n$ turns per unit length carrying a current $I$ , the magnetic field is:

Inside the solenoid ( $r < 2R$ ): $B_{sol} = \mu_0 n I$ . The direction is uniform and parallel to the axis of the solenoid. Let's take this as the z-axis, so $\vec{B}_{sol} = \mu_0 n I \hat{z}$ .
Outside the solenoid ( $r > 2R$ ): $B_{sol} = 0$ .

2. Magnetic Field of the Infinitely Long Hollow Cylindrical Conductor ( $B_{cyl}$ )

We use Ampere's circuital law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$ . Due to symmetry, the magnetic field lines are concentric circles around the axis.

Inside the hollow region ( $r < R$ ): For an Amperian loop of radius $r < R$ , the enclosed current $I_{enclosed} = 0$ . Therefore, $B_{cyl} \cdot (2\pi r) = 0$ , which implies $B_{cyl} = 0$ .
Outside the cylinder ( $r > R$ ): For an Amperian loop of radius $r > R$ , the enclosed current is the total current $I$ flowing through the cylinder, so $I_{enclosed} = I$ . Thus, $B_{cyl} \cdot (2\pi r) = \mu_0 I$ , which gives $B_{cyl} = \frac{\mu_0 I}{2\pi r}$ . The direction of this field is tangential to the circular Amperian loop. Let's denote this by the $\hat{\theta}$ direction.

Now, let's analyze the total magnetic field $\vec{B}_{total} = \vec{B}_{cyl} + \vec{B}_{sol}$ in each region.

Region A: 0 < r < R

In this region, we are inside the hollow part of the cylinder, so $\vec{B}_{cyl} = 0$ .
We are also inside the solenoid, so $\vec{B}_{sol} = \mu_0 n I \hat{z}$ .
The total magnetic field is $\vec{B}_{total} = \vec{B}_{cyl} + \vec{B}_{sol} = 0 + \mu_0 n I \hat{z} = \mu_0 n I \hat{z}$ .
Since $I$ and $n$ are non-zero, the magnetic field is non-zero.
Therefore, statement A is correct.

Region B & C: R < r < 2R

In this region, we are outside the cylinder, so $\vec{B}_{cyl} = \frac{\mu_0 I}{2\pi r} \hat{\theta}$ (tangential).
We are still inside the solenoid, so $\vec{B}_{sol} = \mu_0 n I \hat{z}$ (axial).
The total magnetic field is $\vec{B}_{total} = \frac{\mu_0 I}{2\pi r} \hat{\theta} + \mu_0 n I \hat{z}$ .
The total magnetic field has both a tangential component and an axial component. It is neither purely along the common axis nor purely tangential.
Therefore, statements B and C are incorrect.

Region D: r > 2R

In this region, we are outside the cylinder, so $\vec{B}_{cyl} = \frac{\mu_0 I}{2\pi r} \hat{\theta}$ (tangential).
We are now outside the solenoid, so $\vec{B}_{sol} = 0$ .
The total magnetic field is $\vec{B}_{total} = \vec{B}_{cyl} + \vec{B}_{sol} = \frac{\mu_0 I}{2\pi r} \hat{\theta} + 0 = \frac{\mu_0 I}{2\pi r} \hat{\theta}$ .
For any finite $r > 2R$ , the magnitude of the magnetic field is $B_{total} = \frac{\mu_0 I}{2\pi r}$ , which is non-zero.
Therefore, statement D is correct.

Based on the analysis, the correct statements are A and D.

More from Magnetism

View written solutionFree

Correct answer: A, D

1. Magnetic Field of the Infinite Solenoid ( $B_{sol}$ )

For an ideal infinite solenoid with radius $R_{sol} = 2R$ and $n$ turns per unit length carrying a current $I$ , the magnetic field is:

Inside the solenoid ( $r < 2R$ ): $B_{sol} = \mu_0 n I$ . The direction is uniform and parallel to the axis of the solenoid. Let's take this as the z-axis, so $\vec{B}_{sol} = \mu_0 n I \hat{z}$ .
Outside the solenoid ( $r > 2R$ ): $B_{sol} = 0$ .

2. Magnetic Field of the Infinitely Long Hollow Cylindrical Conductor ( $B_{cyl}$ )

We use Ampere's circuital law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$ . Due to symmetry, the magnetic field lines are concentric circles around the axis.

Inside the hollow region ( $r < R$ ): For an Amperian loop of radius $r < R$ , the enclosed current $I_{enclosed} = 0$ . Therefore, $B_{cyl} \cdot (2\pi r) = 0$ , which implies $B_{cyl} = 0$ .
Outside the cylinder ( $r > R$ ): For an Amperian loop of radius $r > R$ , the enclosed current is the total current $I$ flowing through the cylinder, so $I_{enclosed} = I$ . Thus, $B_{cyl} \cdot (2\pi r) = \mu_0 I$ , which gives $B_{cyl} = \frac{\mu_0 I}{2\pi r}$ . The direction of this field is tangential to the circular Amperian loop. Let's denote this by the $\hat{\theta}$ direction.

Now, let's analyze the total magnetic field $\vec{B}_{total} = \vec{B}_{cyl} + \vec{B}_{sol}$ in each region.

Region A: 0 < r < R

In this region, we are inside the hollow part of the cylinder, so $\vec{B}_{cyl} = 0$ .
We are also inside the solenoid, so $\vec{B}_{sol} = \mu_0 n I \hat{z}$ .
The total magnetic field is $\vec{B}_{total} = \vec{B}_{cyl} + \vec{B}_{sol} = 0 + \mu_0 n I \hat{z} = \mu_0 n I \hat{z}$ .
Since $I$ and $n$ are non-zero, the magnetic field is non-zero.
Therefore, statement A is correct.

Region B & C: R < r < 2R

In this region, we are outside the cylinder, so $\vec{B}_{cyl} = \frac{\mu_0 I}{2\pi r} \hat{\theta}$ (tangential).
We are still inside the solenoid, so $\vec{B}_{sol} = \mu_0 n I \hat{z}$ (axial).
The total magnetic field is $\vec{B}_{total} = \frac{\mu_0 I}{2\pi r} \hat{\theta} + \mu_0 n I \hat{z}$ .
The total magnetic field has both a tangential component and an axial component. It is neither purely along the common axis nor purely tangential.
Therefore, statements B and C are incorrect.

Region D: r > 2R

In this region, we are outside the cylinder, so $\vec{B}_{cyl} = \frac{\mu_0 I}{2\pi r} \hat{\theta}$ (tangential).
We are now outside the solenoid, so $\vec{B}_{sol} = 0$ .
The total magnetic field is $\vec{B}_{total} = \vec{B}_{cyl} + \vec{B}_{sol} = \frac{\mu_0 I}{2\pi r} \hat{\theta} + 0 = \frac{\mu_0 I}{2\pi r} \hat{\theta}$ .
For any finite $r > 2R$ , the magnitude of the magnetic field is $B_{total} = \frac{\mu_0 I}{2\pi r}$ , which is non-zero.
Therefore, statement D is correct.

Based on the analysis, the correct statements are A and D.

Magnetism question

Magnetism question

More from Magnetism

Magnetism question

Magnetism question

More from Magnetism