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5.2 (a) Earth's magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth's magnetic field with the time cannot be neglected.

(b) Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence this is not considered as a source of earth's magnetism.

(c) The radioactivity in earth's interior is the source of energy that sustains the currents in the outer conducting regions of earth's core. These charged currents are considered to be responsible for earth's magnetism.

(d) The change of earth's magnetic field got weakly recorded in rocks during their solidification. One can get a clue after analysing this rock magnetism.

(e) It departs because of the presence of ionosphere. In this region, earth's field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.

(f) An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

5.1 (a) Earth’s magnetic field can be specified by three following independent quantities

Magnetic declination

Angle of dip

Horizontal component of earth’s magnetic field.

(b) The angle of dip at a point depends on how far the point is located with respect to North pole or South pole. The angle of dip will be more in Britain than Southern India as Britain is closer to Magnetic North pole than South India to the Magnetic South pole.

(c) It is a hypothesis that a huge bar magnet is embedded deep in Earth’s ground with its north pole near magnetic south pole of earth and south pole is near magnetic north pole of earth. Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

(d) If a compass is located on the geomagnetic North pole or South pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such case, the compass can point in any direction.

(e) Magnetic moment, M = 8 $\times {10}^{22}$ J $T^{-1}$

Radius of the Earth, r = 6.4 $\times {10}^6$ m

Magnetic field strength is given by the expression:

B = $\frac{{\mu }_{0}M}{4\pi r^3}$ . where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Hence, B = $\frac{4\pi \times {10}^{-7}\times 8\mathrm{}\times {10}^{22}}{4\pi \times {(6.4\mathrm{}\times {10}^6)}^3}$ = 0.305 G

(f) This quantity is of the order of magnitude of the observed field on earth.Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $\times {10}^{-3}$ kg $\times {10}^{-3}$

Hence, the current flowing through the solenoid is 108 A.

4.24 Magnetic field strength, B = 3000G = 3000 $\times {10}^{-4}$ T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop, A = l $\times b$ = 10 $\times 5$ = 50 ${cm}^2$ = 50 $\times {10}^{-4}$ $m^2$

Current in the loop, I = 12 A

Assume that the anti-clockwise direction of the current is positive and vice versa.

Torque, $\stackrel{?}{\tau }$ = $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

From the given figure, it can be observed that A is normal to the y-z plane and B is directed along z-axis.

$\tau$ = 12*( 50 $\times {10}^{-4})\stackrel{?}{i}\times 0.3\stackrel{?}{k}$ = $-1.8\times {10}^{-2}\stackrel{?}{j}$ Nm

The Torque $1.8\times {10}^{-2}$ is Nm along the negative y-direction.

The force on the loop is zero because the angle between A & B is zero.

This case is similar to case (a). The answer is same as case (a)

Torque, $\stackrel{?}{\tau }$ = I $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

From the given figure, it can be observed that A is normal to the x-z plane and B is directed along z-axis.

= - 12 $\times$ (50 $\times {10}^{-4})\stackrel{?}{i}\times 0.3\stackrel{?}{k}$ = $1.8\times {10}^{-2}$ Nm

The Torque is $1.8\times {10}^{-2}$ Nm at an angle 240 $°$ with positive x direction. The force is zero

Magnitude of the torque is given as:

| $\left|\tau \right|$ |= IAB|

12 $\times$ 50 $\times {10}^{-4}\times 0.3$ = $1.8\times {10}^{-2}$ Nm

The Torque is $1.8\times {10}^{-2}$ is Nm at an angle 240 with positive x direction. The force is zero.

Torque, $\left|\tau \right|$ = I $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

= (12 $\times$ 50 $\times {10}^{-4})\stackrel{?}{k}$ $\times 0.3\stackrel{?}{k}$

= 0

The torque is zero and the force is also zero.

Torque, $\stackrel{?}{\tau }$ = I $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

= (12 $\times$ 50 $\times {10}^{-4})\stackrel{?}{k}$ $\times 0.3\stackrel{?}{k}$

= 0

The torque is zero and the force is also zero.

In the case of (e), the direction of I $\stackrel{?}{A}$ and $\stackrel{?}{B}$ is the same and the angle between them is zero. If displaced, they come back to equilibrium. Hence, its equilibrium is stable.

In the case of (f), the direction of I $\stackrel{?}{A}$ and $\stackrel{?}{B}$ is opposite. The angle between them is 180 $°$ . If disturbed, it does not come back to its original position, hence the equilibrium is unstable.