मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre.

Advertisements
Advertisements

प्रश्न

A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it. It is now connected across a long solenoid having 4000 turns per metre. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the centre of the solenoid during this period. 

टीपा लिहा
Advertisements

उत्तर

Given:
Capacitance, C = 100 microfarad
Voltage, V = 20 V
Charge stored in the capacitor, Q = CV 

\[= 100 \times {10}^{- 6} \times 20\]
\[ = 2 \times {10}^{- 3} C\]

It is given that the potential difference across the capacitor drops to 90% of its maximum value.
Thus,

\[V' = \frac{90}{100} \times 20 = 18 V\]
\[\text{ New charge, }Q' = CV' = 1 . 8 \times {10}^{- 3} C\]
\[\text{ Now, }\]
\[\text{ Current, i }= \frac{Q - Q'}{t}\]
\[ \Rightarrow i = \frac{(2 - 1 . 8) \times {10}^{- 3}}{2} = \frac{2 \times {10}^{- 4}}{2}\]
\[ \Rightarrow i = 1 \times {10}^{- 4} A\]

No. of turns per metre, n = 4000
Thus, the average magnetic field at the centre of the solenoid is given by

\[B = \mu_0 ni\]
\[ = 4\pi \times {10}^{- 7} \times 4000 \times {10}^{- 4} \]
\[ = 16\pi \times {10}^{- 8} \]T 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 35: Magnetic Field due to a Current - Exercises [पृष्ठ २५३]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 35 Magnetic Field due to a Current
Exercises | Q 61 | पृष्ठ २५३

संबंधित प्रश्‍न

Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to ‘I’.


Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius 'r', having 'n' turns per unit length and carrying a steady current I.


An observer to the left of a solenoid of N turns each of cross section area 'A' observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA.

 


Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r1 and r2 (r2 >> r1). Total number of turns in the two solenoids are N1 and N2, respectively.


A magnetic field of 100 G (1 G = 10−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10−3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.


Define self-inductance of a coil.


A wire AB is carrying a steady current of 10 A and is lying on the table. Another wire CD carrying 6 A is held directly above AB at a height of 2 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms−2]


A wire AB is carrying a steady current of 6 A and is lying on the table. Another wire CD carrying 4 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms−2]


The magnetic field inside a tightly wound, long solenoid is B = µ0 ni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra-added loops do not have a considerable effect on the field inside the solenoid.  


The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14 × 10−2 T. Find the number of turns per unit length of the solenoid. 


A copper wire having resistance 0.01 ohm in each metre is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0 × 10−2 T near the centre of the solenoid.


A tightly-wound solenoid of radius a and length l has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains n dx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid. (b) verify that if l >> a, the field tends to B = µ0ni and if a >> l, the field tends to `B =(mu_0nil)/(2a)` . Interpret these results.


A tightly-wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00 × 108 rev s−1. Find the number of turns per metre in the solenoid. 


A current of 1.0 A is established in a tightly wound solenoid of radius 2 cm having 1000 turns/metre. Find the magnetic energy stored in each metre of the solenoid.


Magnetic field inside a solenoid is ______.

A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×