A magnetic compass needle oscillates 30 times per sec at a place where the dip angle is 45° , and 40 times per second where the dip angle is 30o . If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2 is nearly equal to :

Question

A magnetic compass needle oscillates 30 times per sec at a place where the dip angle is 45° , and 40 times per second where the dip angle is 30o . If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2 is nearly equal to : (A) 0.7 (B) 3.6 (C) 1.8 (D) 2.2

Tardigrade Admin · Accepted Answer

n=(1/2 π )√(M BH/1) BH=Becosφ (n1/n2)=√(B1 cos φ1/B2 cos φ2) (30/40)=√(B1 cos 45 ° /B2 cos 30 ° ) =(9/16)× (√6/2)=0.68

$0.7$

$3.6$

$1.8$

$2.2$

$n = \frac{1}{2 π} \frac{M B _{H}}{1}$
$B_{H} = B_{e} cos ϕ$
$\frac{n _{1}}{n _{2}} = \frac{B _{1} cos ϕ _{1}}{B _{2} cos ϕ _{2}}$
$\frac{30}{40} = \frac{B _{1} cos 4 5 ^{\circ}}{B _{2} cos 3 0 ^{\circ}}$
$= \frac{9}{16} \times \frac{6}{2} = 0.68$

Q. A magnetic compass needle oscillates 30 times per sec at a place where the dip angle is 45∘, and 40 times per second where the dip angle is 30o . If B1​ and B2​ are respectively the total magnetic field due to the earth at the two places, then the ratio B1​/B2​ is nearly equal to :

0.7

3.6

1.8

2.2

n=2π1​1MBH​​​ BH​=Be​cosϕ n2​n1​​=B2​cosϕ2​B1​cosϕ1​​​ 4030​=B2​cos30∘B1​cos45∘​​ =169​×26​​=0.68

$0.7$

$3.6$

$1.8$

$2.2$

$n = \frac{1}{2 π} \frac{M B _{H}}{1}$
$B_{H} = B_{e} cos ϕ$
$\frac{n _{1}}{n _{2}} = \frac{B _{1} cos ϕ _{1}}{B _{2} cos ϕ _{2}}$
$\frac{30}{40} = \frac{B _{1} cos 4 5 ^{\circ}}{B _{2} cos 3 0 ^{\circ}}$
$= \frac{9}{16} \times \frac{6}{2} = 0.68$