Magnetism and Matter Easy Questions and Answers | Page - 1

Magnetism and Matter

If θ₁ and θ₂ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by :-

1. tan¹θ = tan¹θ₁ + tan¹θ₂
1. cot¹θ = cot¹θ₁ – cot¹θ₂
1. tan¹θ = tan¹θ₁ – tan¹θ₂
1. cot¹θ = cot¹θ₁ + cot¹θ₂

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If θ₁ and θ₂ are opparent angles of dip
Let α be the angle which one of the plane make with the magnetic meridian.
tan θ₁ = υ
H cos α

i.e.,cos α = υ      ....(i)
H tan θ₁

tan θ₂ = υ ,
H sin α

i.e.,sin α = υ      ....(ii)
H tan θ₂

Squaring and adding (i) and (ii), we get

i.e., 1 = V² [cot²θ₁ - cot²θ₂]
H²

i.e., cot² θ = cot²θ₁ + cot² θ₂

Correct Option: D

If θ₁ and θ₂ are opparent angles of dip
Let α be the angle which one of the plane make with the magnetic meridian.
tan θ₁ = υ
H cos α

i.e.,cos α = υ      ....(i)
H tan θ₁

tan θ₂ = υ ,
H sin α

i.e.,sin α = υ      ....(ii)
H tan θ₂

Squaring and adding (i) and (ii), we get

i.e., 1 = V² [cot²θ₁ - cot²θ₂]
H²

i.e., cot² θ = cot²θ₁ + cot² θ₂

Current i is flowing in a coil of area A and number of turns N, then magnetic moment of the coil, M is

1. NiA
1. Ni
  A
1. Ni
  √A
1. N² Ai

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Magnetic moment linked with one turn = iA Magnetic moment linked with N turns = iNA amp-m². Here, A = Area of current loop.

Correct Option: A

Magnetic moment linked with one turn = iA Magnetic moment linked with N turns = iNA amp-m². Here, A = Area of current loop.

A bar magnet, of magnetic moment M', is placed in a magnetic field of induction B'. The torque exerted on it is

1. M^{^→} . B^{^→}
1. - M^{^→} . B^{^→}
1. M^{^→} × B^{^→}
1. B^{^→} × M^{^→}

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We know that when a bar magnet is placed in the magnetic field at an angle θ, then torque acting on the bar magnet (τ) = MB sin θ = M^{^→} × B^{^→}.
Note : This torque τ has a tendency to make the axis of the magnet parallel to the direction of the magnetic field.

Correct Option: C

We know that when a bar magnet is placed in the magnetic field at an angle θ, then torque acting on the bar magnet (τ) = MB sin θ = M^{^→} × B^{^→}.
Note : This torque τ has a tendency to make the axis of the magnet parallel to the direction of the magnetic field.

The work done in turning a magnet of magnetic moment M by an angle of 90° from the meridian, is n times the corresponding work done to turn it through an angle of 60°. The value of n is given by

1. 2
1. 1
1. 0.5
1. 0.25

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Magnetic moment = M; Initial angle through which magnet is turned (θ₁) = 90º and final angle through which magnet is turned (θ₂)= 60º. Work done in turning the magnet through 90º(W₁) = MB (cos 0º – cos 90º) = MB (1–0) = MB.
Similarly, W₂ = MB (cos 0º – cos 60º)

∴W₁ = 2W₁ or n = 2.

Correct Option: A

Magnetic moment = M; Initial angle through which magnet is turned (θ₁) = 90º and final angle through which magnet is turned (θ₂)= 60º. Work done in turning the magnet through 90º(W₁) = MB (cos 0º – cos 90º) = MB (1–0) = MB.
Similarly, W₂ = MB (cos 0º – cos 60º)

∴W₁ = 2W₁ or n = 2.

A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that B^{^→} is in the plane of the coil. If due to a current i in the triangle a torque τ acts on it, the side l of the triangle is

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τ = MB sinθ
τ = iAB sin90º

∴ A = τ
iB

Also, A = 1/2 (BC) (AD)

Correct Option: B

τ = MB sinθ
τ = iAB sin90º

∴ A = τ
iB

Also, A = 1/2 (BC) (AD)