Physics

5.14 The magnetic field on the axis of the magnet at a distance $d_1$ = 14 cm, can be written as

$B_1$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{d_1^3}$ = H ……………….(1)

where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$ and M = magnetic moment

If the bar magnet is turned through 180 $°$ , then the neutral point will lie on the equatorial line. Hence, the magnetic field at a distance , on the equatorial line of the magnet can be written as:

$B_2$ = $\frac{{\mu }_0}{4\pi }\frac{M}{d_2^3}$ = H ……………….(2)

where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$ and M = magnetic moment

Equating (1) and (2), we get

$\frac{2}{d_1^3}$ = $\frac{1}{d_2^3}$

${\left(\frac{d_2}{d_1}\right)}^3$ = $\frac{1}{2}$

$d_2=d_1*({\frac{1}{2})}^{\frac{1}{3}}$ = 14 $*0.793$ = 11.11 cm

The new null points will be located 11.11 cm on the normal bisector.

5.13 Earth's magnetic field at the given place, H = 0.36 G

The magnetic field at a distance d, on the axis of the magnet is given as

$B_1$ = $\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ = H ……………….(1)

where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$ and M = magnetic moment

Total magnetic field, B = $B_1+B_2$ = H + $\frac{H}{2}$ = 0.36 + 0.18 = 0.54 G

Therefore, the magnetic field is 0.54 G in the direction of earth's magnetic field.

5.12 Magnetic moment of the bar magnet, M = 0.48 J/T

The distance, d = 10 cm = 0.1 m

The magnetic field at a distance d from the centre of the magnet on the axis is given by the relation:

B = $\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$ $\pi *{10}^{-7}$

B = $\frac{4\pi *{10}^{-7}*2*0.48}{4\pi *{0.1}^3}$ = 0.96 $*{10}^{-6}$ T = 0.96 G

The magnetic field is along S – N direction.

The distance, d = 10 cm = 0.1 m

The magnetic field at a distance d from the centre of the magnet on the equatorial line of the magnet is given by the relation:

B = $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$

B = $\frac{4\pi *{10}^{-7}*0.48}{4\pi *{0.1}^3}$ = 0.48 T = 0.48 $*{10}^{-6}$ G

The magnetic field is along N – S direction.

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