Tangent Law 211729

Class XII/Physics

TANGENT LAW

If a small bar magnet is suspended in two mutually perpendicular magnetic fields

 BH , then B = B H tan θ ----------- (1)

  B and BH , such that it comes

to rest making an angle θ with the direction of the field

Proof: Consider a small bar magnet NS placed in two uniform magnetic fields

  B and BH acting at right angles to each other. Let ‘m’ be the pole strength of

each pole and ‘2l’ be the length of the magnet. The forces on the north-pole and the south-pole are shown in the figure. Each of the two magnetic fields produces a torque on the magnet. The torque due to

  BH tends to rotate the magnet in the anti-clockwise direction, while that due to B in

clockwise direction. Suppose the magnet comes to the equilibrium position, making an angle θ with the direction of the field Torque due to the magnetic field

τ 1 = mB H × KN Torque due to the magnetic field

 BH .

 BH ,  B,

τ 2 = mB × KS In equilibrium, τ 1 = τ 2 i.e. mB H × KN = mB × KS KN ------------- (2) KS KN From the ΔSKN, tan θ = --------------- (3) KS From equations (2) and (3), we get- B = B H tan θ Or,

B = BH

It proves the tangent law. Tangent Galvanometer A tangent galvanometer is an instrument to measure very small currents. It is a galvanometer of moving magnet and fixed coil type. Principle: It is based on tangent law in magnetism. Construction: A tangent galvanometer consists of a coil of insulated copper wire wound on a circular nonmagnetic frame. The frame is mounted vertically on a horizontal base provided with levelling screws. The coil can be rotated on a vertical axis ing through its centre. A com box is mounted horizontally at the centre of a circular scale. It consists of a tiny, powerful magnetic needle pivoted at the centre of the coil. The magnetic needle is free to rotate in the horizontal plane. The circular scale is divided into four quadrants. Each quadrant is graduated from 0° to 90°. A long thin aluminium pointer is attached to the needle at its centre and at right angle to it. To avoid errors due to parallax a plane mirror is mounted below the com needle.  Theory: When current is ed through the coil, the magnetic field B is produced at the centre of the coil in a direction perpendicular to the plane of the coil. The galvanometer is arranged in such a way that the horizontal component of earth’s magnetic field

 BH is in the plane of the coil. The magnetic needle will be under the action

of two mutually perpendicular magnetic fields. If θ is the deflection of the needle, then according to the tangent lawB = B H tan θ ------------------ (1) If the coil has ‘n’ turns and radius ‘a’, then on ing current ‘I’, the magnetic field produced at the centre of the coil is given by-

B=

µ0 2πnI . 4π a

--------------- (2)

From equations (1) and (2), we have-

µ 0 2πnI . = B H tan θ 4π a

2a B H tan θ ------------------- (3) nµ 0

Or,

I=

Knowing ‘a’, ‘n’, ‘θ’ and

 BH , the value of the current I through the coil can be calculated.

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Class XII/Physics

In the above equation, equation becomes-

The factor becomes-

BH G

nµ0 2a

is called the galvanometer constant and is denoted by G. therefore, the above

I=

BH tan θ G

------------------ (4)

is called the reduction factor of the tangent galvanometer and is denoted by K. Therefore, eqn. (4)

I = K tan θ

----------------- (5)

If θ=45°, then from eqn. (5),

Or I = K tan 45 ° = K × 1 = K K =I Thus, the reduction factor of a tangent galvanometer is numerically equal to the current through the galvanometer, which produces a deflection of 45°, when plane of the coil lies in the magnetic meridian. VIBRATION MAGNETOMETER

Vibration magnetometer is used for comparison of magnetic moments and magnetic fields. Principle: It is based on the principle that when a bar magnet is suspended freely in a uniform magnetic field is displaced from its equilibrium position, it starts executing simple harmonic motion about the equilibrium position. Theory: Consider a magnet with magnetic dipole moment M suspended freely in



earth’s magnetic field BH

. If the magnet is displaced through an angle θ, a torque

equal to MB H sin θ acts on it which tends to bring it back to its equilibrium position. If ‘I is the moment of inertia of the magnet about the suspension axis and α, the angular acceleration produced, then-

Iα = −MB H sin θ

If θ is small, then sin θ≈ θ, hence the angular acceleration (in magnitude) is given by-

α= MB H I

MB H θ I

------------------------- (1)

is a constant for a given magnet and hence, angular acceleration is directly proportional to the angular

displacement.

α ∝θ

Therefore, the time period of the magnet is given by-

T = 2π

angular displaceme nt θ = 2π angular accelerati on α

T = 2π

I -------------------- (2) MB H

Using equation (1), we get-

For a rectangular bar,

I = mass ×

length 2 + breadth 2 12

Uses of a vibration magnetometer: (i) To compare the magnetic moments of two bar magnets: consider two bar magnets A and B of same size and mass. Let M1 and M2 be the magnetic moments of magnets A and B respectively and I be the moment of inertia of each magnet. Set the vibration magnetometer, so that the suspension thread and the stirrup rests along the N-S line. Place the magnet A in the stirrup and find its period T1. Then,

T1 = 2π

 I ------------- (3), where BH is the horizontal component M 1 BH

of earth’s magnetic field. Next, magnet A is replaced by magnet B and its time period is noted as T2. Then-

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Class XII/Physics

T2 = 2π

I ------------ (4) M 2 BH

Dividing eqn. (3) by eqn. (4), we get-

T1 = T2

Or,

Knowing T1 and T2, the ratio

T1

2

T2

2

M1 M2

2π 2π

I M 1 BH I M 2 BH

=

I × M 1 BH

M 2 BH = I

M2 M1

2

M M T = 2 ⇒ 2 = 12 M1 M 1 T2 can be found.

(ii) To compare the horizontal component of earth’s magnetic field at two places: Consider a magnet having moment of inertia I and magnetic moment M. Let

 BH and

 B H ' be the values of horizontal component

of earth’s magnetic field at two places A and B respectively. If T and T’ are the values of time period of the magnet at the places A and B respectively, then-

I ----------------- (5) MB H

T = 2π

And,

T ' = 2π

I MB H '

---------------- (6)

Dividing eqn. (5) by eqn. (6), we get-

T T'

Squaring,

Knowing T and T’, the ratio

2π = 2π

I MB H I

=

MB H '

MB H ' I × = MB H I

BH ' BH

BH ' T 2 = B H T '2 BH

BH '

can be found.

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