Physics

5.11 Given:

The angle of declination, $\theta$ = 12 $°$

The angle of dip, $\beta$ = 60 $°$

Horizontal component of Earth's magnetic field, $B_H$ = 0.16 G

If Earth's magnetic field be B, we can relate B and $B_H$ as $B_H=B\cos ?\beta$

B = $\frac{B_H}{\cos ?\beta }$ = $\frac{0.16}{\cos ?60°}$ = 0.32 G

Therefore, the Earth's magnetic field lies in the vertical plane, 12west of the geographic meridian, making an angle of 60 $°$ (upward) with the horizontal direction. Its magnitude is 0.32 G.

5.10 Given:

Horizontal component of Earth's magnetic field, $B_H$ = 0.35 G

Angle made by the needle with the horizontal plane, $\beta$ = 22 $°$

If the Earth's magnetic field at that location be B,

then $B_H$ = B $\cos ?\beta$

B = $\frac{B_H}{\cos ?\beta }$ = $\frac{0.35}{\cos ?22°}$ = 0.377 G

Therefore, the strength of Earth's magnetic field at that location is 0.377 G

5.9 Number of turns, n = 16

Radius of the coil, r = 10 cm = 0.1 m

Cross-section of the coil, A = $\pi r^2$ = $\pi *{0.1}^2$ $m^2$ = 0.0314 $m^2$

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 $*{10}^{-2}$ T

Frequency of oscillation of the coil, $\nu$ = 2/s

Magnetic moment, M = NIA = 16 $*0.75*0.0314$ = 0.377 A $m^2$

Frequency is given by the relation, $\nu$ = $\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$ , where I = moment of Inertia of the coil

I = $\frac{MB}{4{\pi }^2{\nu }^2}$ = $\frac{0.377\mathrm{}*5.0\mathrm{}*{10}^{-2}}{4{\pi }^2*2^2}$ = 1.19 $*{10}^{-4}$ kg $m^2$

5.8 (a) Number of turns, n = 2000

Area of cross-section, A = 1.6 $*{10}^{-4}$ $m^2$

Current, I = 4.0 A

The magnetic moment along the axis of the solenoid is calculated as

M = nAI = 2000 $*$ 1.6 $*{10}^{-4}*4$ = 1.28 A $m^2$

(b) Magnetic field, B = 7.5 $*{10}^{-2}$ T

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

Torque $\tau =MB\sin ?\theta$

= 1.28 $*$ 7.5 $*{10}^{-2}$ $*\sin ?30°$

= 4.8 $*{10}^{-2}$ Nm

5.6 Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 J/T

The angle $\theta$ , between the axis of the solenoid and the direction of the applied field is 30 $°$ . Therefore, the torque acting on the solenoid is given as

$\tau =MBsin\theta$ = 0.6 $*0.25*\sin ?30°$ = 7.5 $*{10}^{-2}$ J

5.5 Number of turns, n = 800

Area of the cross-section, A = 2.5 $*{10}^{-4}$ $m^2$

Current flowing, I = 3.0 A

A current carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis (along the length). $°$

The magnetic moment associated is calculated as M = nIA = 800 $*3*$ 2.5 $*{10}^{-4}$ J/T= 0.6 J/T

5.4 Moment of the bar magnet, M = 0.32 J/T

Magnetic field, B = 0.15 T

The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle $\theta$ , between the bar magnet and the magnetic field is 0 $°$ .

Potential energy of the system = -MBcos $\theta$ = - 0.32 $*0.15*\cos ?0°$ = -4.8 $*{10}^{-2}$ J

When the bar magnet is oriented 180 $°$ to the magnetic field, it becomes unstable equilibrium.

Potential energy = - MBcos $\theta$ = - 0.32 $*0.15*\cos ?180°$ = 4.8 $*{10}^{-2}$ J

5.3 Magnetic field strength, B = 0.25 T

Torque on the bar magnet, $\tau$ = 4.5 $*{10}^{-2}$ J

Angle between the bar magnet and the external magnetic field, $\theta$ = 30 $°$

From the relation T = MB $\sin ?\theta$ , where M = Magnetic moment, we get

M = $\frac{\tau }{B\sin ?\theta }$ = $\frac{4.5\mathrm{}*{10}^{-2}}{0.25sin30°}$ = 0.36 J/T

Hence the magnetic moment is 0.36 J/T