Physics 3 Marks Question

1. Given,

Number of turns of coil. N = 2000

Area of cross-section of solenoid, A = 1.6 x 10^-4 m²

Current passing through the coil. I = 4 A

Magnetic moment. M = NIA

= 2000 × 4 × 1.6 × 10^-4

= 1.28 Am²

The direction of $\overrightarrow{\text{M}}$ is along the axis of the solenoid in the direction related to the sense of current via the right handed screw rule.

2. Uniform magnetic field applies, B = 7.5 × 10^-2 T

Angle between the axis of the solenoid and magnetic field, $\theta=30^\circ$

Since the magnetic field is uniform on the solenoid, force acting on the solenoid is 0.

Torque is given by,

$\tau=\text{MB }\sin\theta$

$=1.28\times7.5\times10^{-2}\sin 30^\circ$

$=1.28\times7.5\times10^{-2}\times\frac{1}{2}\text{ J}$

$=0.048\text{ J}.$

The direction of the torque is such that the solenoid tends to align the axis of the solenoid (magnetic moment vector) along the direction of magnetic field $\overrightarrow{\text{B}}.$

1. Work done on magnet

1. Amount of work required to turn the magnet normal to the field direction is,

$\text{W}_{1}=-\text{MB}\big[\cos90^\circ-\cos0^\circ\big]$

= + MB

= MB = 1.5 × 0.22

= 0.33 N-m

2. Amount of work required to turn the magnet in a direction opposite to the field direction is,

$\text{W}_{2}=-\text{MB}\big[\cos180^\circ-\cos0^\circ\big]$

= 2 MB

$=-\text{MB}\big[\cos\theta_{2}-\cos\theta_{1}\big]$

= 2 MB = 2 × 0.33

= 0.66 J

2. Torgue acting on the magnet.

1. $\tau=\text{MB}\sin 90^\circ$

= MB

= 0.33J.

It works in the direction that tends to align the magnetic moment vector along B.

2. Torque acting on the magnet so as to align it's magnetic moment in a direction opposite to the field direction.

$\tau=\text{MB }\sin\theta$

Here, $\theta=180^\circ$

$\tau=1.5\times0.22\times\sin180^\circ$

= 1.5 × 0.22 × 0

= 0.

1. The magnetic field due to current I, at a distance r, is given as:

$\text{B}=\frac{\mu_02\pi\text{NI}}{4\pi\text{r}}$

Where,

$\mu_0$ = Permeability of free space = $4\pi\times10^{-7}\text{ TM A}^{-1}$

$\therefore\ \text{B}=\frac{4\pi\times10^{-7}\times2\pi\times30\times0.35}{4\pi\times0.12}$

$=5.49\times10^{-5}\text{ T}$

The compass needle points from west to East. Hence, the horizontal component of earth's rnagnetic field is given as:

$\text{B}_\text{H}=\text{B}\sin\delta$

$=5.49\times10^{-5}\sin45^\circ$ $=3.88\times10^{-5}\text{ T}=0.388\text{ G}$

2. When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90°, the needle will reverse its original direction. In this case, the needle will point from East to West.

1. $\text{i}=\frac{\text{q}}{\text{t}}=\frac{\text{q}}{\text{t}}=\frac{\text{q}}{\Big(\frac{2\pi}{\omega}\Big)}=\frac{\text{q}\omega}{2\pi}$

2. $\mu=\text{n}\text{ ia}=\text{i A}[\because\text{n}=1]$

$=\frac{\text{q}\omega\pi\text{r}^2}{2\pi}$

3. $\mu=\frac{\text{q}\omega\text{r}^2}{2}$

$\text{L}=\text{I}\omega=\text{mr}^2\omega,\ \frac{\mu}{\text{L}}$

$=\frac{\text{q}\omega\text{r}^2}{2\text{mr}^2\omega}=\frac{\text{q}}{2\text{m}}$

$\Rightarrow\mu=\Big(\frac{\text{q}}{2\text{m}}\Big)\text{L}$

1. Material is diamagnetic (ii) Material is paramagnetic. $\mu\text{r} = 1+ \text{X}$

1. Range of relative magnetic permeability for diamagnetic is $0 \leq \mu < 1.$
2. Range of relative magnetic permeability for paramagnetic is $1< \mu\text{r} <1+ \varepsilon$

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1. $=0.4\times10^-4\text{Wb/m}^2,\theta=30^\circ$

$\tan\theta=\frac{\text{V}}{\text{H}}$

$\Rightarrow\text{vertical component}\text{V}=\text{H}\tan\theta$

$=0.4\times10^{-4}\times\tan^{30^\circ}$

$=\frac{0.4\times10^{-4}}{\sqrt{3}}=0.23\times10^{-4}\text{Wb/m} ^2$

2. Total intensity of earth's magnetic field

$\text{Be}=\sqrt{\text{H}^2+\text{V}^2}$

$=\sqrt{(0.4\times1^{-4})^2+\big(\frac{04.\times10^{-4}}{\sqrt{3}}\big)^2}$

$=.46\times1^{-4}\text{Wb/m}^2$

1. H = 1500A/m

As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, the end effects may be neglected. There is no effect of the rod on the magnetic intensity at the centre.

2. I = 0.12A/m

We know $\overrightarrow{\text{I}}=\text{X}\overrightarrow{\text{H}}$

X = Susceptibility

$\Rightarrow\text{X}=\frac{\text{I}}{\text{H}}=\frac{0.12}{1500}=0.00008=8\times10^{-5}$

3. The material is paramagnetic.