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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Assume the dipole model for earth’s magnetic field B which is given by BV = vertical component of magnetic field = mu_0/(4pi) (2m cos theta)/r^3 BH = Horizontal component of magnetic field - Physics

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प्रश्न

Assume the dipole model for earth’s magnetic field B which is given by BV = vertical component of magnetic field = `mu_0/(4pi) (2m cos theta)/r^3` BH = Horizontal component of magnetic field = `mu_0/(4pi) (sin theta m)/r^3` θ = 90° – lattitude as measured from magnetic equator. Find loci of points for which (i) |B| is minimum; (ii) dip angle is zero and (iii) dip angle is ± 45°.

दीर्घउत्तर
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उत्तर

(i) We know from the figure.


`B^2 + B_V^2 + B_H^2`

= `[mu_0/(4pi) (2m  cos theta)/r^3]^2 + [mu_0/(4pi) (m  sin theta)/r^3]^2`

Substituting the value of BV and BH from question

= `[mu_0/(4pi)]^2 m^2/r^3 [4cos^2 theta + sin^2 theta]`

`B^2 = (mu_0/(4pi)) xx m^2/r^3 [3 cos^2 theta + 1]`

`B = mu_0/(4pi) m/r^3 [3 cos^2 theta + 1]^(1/2)`  ......(i)

From equation (i), the value of B will be minimum when `[3cos^2 theta + 1]^(1/2)` is minimum which will be at `theta= pi/2`. So magnetic equator lies ar `theta = pi/2`, i.e., from magnetic dipole axis `theta = pi/2` for magnetic equator.

(ii) For angle of dip δ

tan δ = `B_r/B_H = (mu_0/(4pi) (2m cos theta)/r^3)/(mu_0/(4pi) (m sin theta)/r^3)`

tan δ = 2 cot θ

For δ = 0, cot θ = 0, θ = `pi/2`

So angle of dip will lie at magnetic equator.

(iii) tan δ = `B_V/B_H` = angle of dip δ = ± 45°

⇒ `B_V/B_H` = tan ± 45° 

⇒ `B_V/B_H` = tan 45° 

`B_V/B_H` = 1 or BV = BH

δ = ± 45°

tan ± 45° = 2 cos θ .....[∵ tan δ = 2 cot θ]

cot θ = `1/2` or tan θ = 2

θ = tan–12 is the locus of points where the angle of dip δ = ± 45°.

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पाठ 5: Magnetism And Matter - MCQ I [पृष्ठ ३२]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 12
पाठ 5 Magnetism And Matter
MCQ I | Q 5.23 | पृष्ठ ३२

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