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प्रश्न
- A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
- Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)
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उत्तर
(a) Number of turns on the circular coil, n = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil = πr2
= π(0.08)2
= 3.14 × 0.0064
= 0.0201 m2
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface,
θ = 60°
The coil experiences a torque in the magnetic field. Hence, it turns.
The counter torque applied to prevent the coil from turning is given by the relation,
τ = n IBA sin θ ...(i)
= 30 × 6 × 1 × 0.0201 × sin 60°
= `3.618 xx sqrt3/2`
= `3.618 xx 1.732/2`
= 3.618 × 0.866
= 3.133 N m
(b) It can be inferred from the relation (i) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.
संबंधित प्रश्न
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters.)
Determine the ratio of
- current sensitivity and
- voltage sensitivity of M2 and M1.
Can a galvanometer as such be used for measuring the current? Explain.
With the help of a neat and labelled diagram, explain the principle and working of a moving coil galvanometer ?
Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter?
In the meter bridge experiment, balance point was observed at J with AJ = l.
(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the alance point get affected?
Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range ?
State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer can not be used as such to measure current in a given circuit. Name any two factors on which the current sensitivity of a galvanometer depends.
What are the advantages of using soft iron as a core, instead of steel, in the coils of galvanometers?
Explain the significance of a radial magnetic field when a current-carrying coil is kept in it.
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
The conversion of a moving coil galvanometer into a voltmeter is done by ______.
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A galvanometer shows full-scale deflection for current Ig. A resistance R1 is required to convert it into a voltmeter of range (0 - V) and a resistance R2 to convert it into a voltmeter of range (0 - 2V). Find the resistance of the galvanometer.
Explain in brief the basic construction of a moving-coil table galvanometer whit a neat labelled diagram.
A resistance of 3Ω is connected in parallel to a galvanometer of resistance 297Ω. Find the fraction of current passing through the galvanometer.
A moving coil galvanometer of resistance 55 Ω produces a full scale deflection for a current of 250 mA. How will you convert it into an ammeter with a range of 0 - 3A?
A galvanometer of resistance 100 Ω gives a full-scale deflection for a potential difference of 200 mV.
- What must be the resistance connected to convert the galvanometer into an ammeter of the range 0-200 mA?
- Determine resistance of the ammeter.
