Q: At a place the value of B_{H} and B_{V} are 0.4\times 10^{-4} and 0.3\times 10^{-4} respectively. The resultant earths magnetic field is :

I=\sqrt{B{_{H}}^{2}+B{_{V}}^{2}} =\sqrt{\left ( 0.4\times 10^{-4} \right )^{2}+\left(0.3\times 10^{-4} \right )^{2}} =0.5\times 10^{-4} T

Question 1

A bar magnet has pole strength  1 Am  and is placed as shown in Fig.  24.19 (a). Find the magnetic field at  P .

Accepted Answer

SN=6cm   ,   SP=8cm  ,   NP=\sqrt{8^2+6^2}=10cm Magnetic field due to south pole is  B_S=\cfrac{\mu_0 M}{4\pi r^2}=\cfrac{\mu_0 M}{4\pi (8	imes 10^{-2})^2} Magnetic field due to north pole is  B_N=\cfrac{\mu_0 M}{4\pi r^2}=\cfrac{\mu_0 M}{4\pi (10	imes 10^{-2})^2} B=\sqrt{B^2_S+B^2_N+2B_SB_N\cos143}   B=\sqrt{B^2_S+B^2_N+2B_SB_N\cos(90+53)} B =\cfrac{\mu_0M}{4\pi} \sqrt{\dfrac{1}{(8	imes10^{-2})^2} +\cfrac{1}{(10	imes10^{-2})^2}-\cfrac{2	imes1}{(8	imes10^{-2})^2}	imes\cfrac{1}{(10	imes10^{-2})^2} \sin53} B =0.22	imes10^{-4}  T=22\mu T

Question 2

At a place the value of  B_{H}  and  B_{V}  are   0.4	imes 10^{-4} and  0.3	imes 10^{-4} respectively. The resultant earths magnetic field is :

Accepted Answer

I=\sqrt{B{_{H}}^{2}+B{_{V}}^{2}} =\sqrt{\left ( 0.4	imes 10^{-4} ight )^{2}+\left(0.3	imes 10^{-4} ight )^{2}} =0.5	imes 10^{-4}  T

Question 3

Horizontal component of the earth's field is  3 	imes 10^{-5} \, T   and of dip is  53^o . Magnetic field of the earth at that place is

Accepted Answer

Horizontal component of the earths magnetic fields is given by B_H=B \, cos \delta  \Rightarrow B=\frac {B_H}{cos\,\delta }=\frac {3	imes 10^{-5}}{cos\, 53^o}  = 5 	imes 10^{-5}\, T

Question 4

Two short bar magnets of magnetic moments  0.125 Am^{2}  and  0.512 Am^{2}  are placed with their like poles facing each other. If the distance between the centres of the magnet is  0.26 m . The distance of neutral point from the weaker magnet is

Accepted Answer

B_{axial} = \dfrac{\mu_0 2M}{4\pi r^3} Equating the magnetic fields due to the two bar magnets,  \dfrac{M_1}{x^3} = \dfrac{M_2}{{(d-x)}^3} Solving the above for x (distance from the center of weaker magnet to the neutral point), gives the formula in the hint. Putting in the values, we get,  x = 0.1m

Question 5

A short bar Magnet with its north pole facing North forms a neutral point P in the horizontal plane. If the magnet is rotated by 90 $^{0}$ in the horizontal plane, the net magnetic induction at P is:

(Horizontal component of earths magnetic field is $B_{H}$ )

Accepted Answer

At P, the equatorial field due to magnet  =B_H .After rotation, P lies on the axis of the bar magnet.Hence, at P, field due to magnet  =2B_H Net field is the vector sum of  B_H  and  2B_H , both at right angles to each other giving,  B_{net} = \sqrt{5}B_H

Magnetism And Matter