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What is the significance of the Ardhanareeshwara dance?
The Ardhanareeshwara dance represents the fusion of masculine and feminine energies, symbolizing Lord Shiva and Goddess Parvati as one entity. This dance expresses balance, equality and the importance of both aspects in creation. Movements blend strength with grace, portraying the complementary natuRead more
The Ardhanareeshwara dance represents the fusion of masculine and feminine energies, symbolizing Lord Shiva and Goddess Parvati as one entity. This dance expresses balance, equality and the importance of both aspects in creation. Movements blend strength with grace, portraying the complementary nature of male and female forces. This dance form serves as a spiritual and philosophical metaphor, teaching the idea that true harmony comes from the integration of contrasting yet interconnected energies in life and the universe.
See lessA long straight wire in the horizontal plane carries a current of 15 A in north to south direction. The magnitude and direction of magnetic field at a point 2.5 m east of the wire respectively are :
The magnetic field around a long, straight current-carrying wire is given by: B = μ0I/2πr Substituting values: B = 4π × 10⁻⁷ × 15/2π × 2.5 = 1.2 × 10⁻⁶ T = 1.2 μT Using the right-hand thumb rule, the magnetic field at a point east of the wire points vertically downward. Answer: (A) 1.2 µT, verticallRead more
The magnetic field around a long, straight current-carrying wire is given by:
B = μ0I/2πr
Substituting values:
B = 4π × 10⁻⁷ × 15/2π × 2.5 = 1.2 × 10⁻⁶ T = 1.2 μT
Using the right-hand thumb rule, the magnetic field at a point east of the wire points vertically downward.
Answer: (A) 1.2 µT, vertically downward.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
If an ammeter is to be used in place of voltmeter, then we must connect with the ammeter a :
An ammeter has low internal resistance and is designed to measure current by connecting in series with the circuit. To use it as a voltmeter, a high resistance must be connected in series to limit current and measure potential difference. This combination ensures minimal current through the meter whRead more
An ammeter has low internal resistance and is designed to measure current by connecting in series with the circuit. To use it as a voltmeter, a high resistance must be connected in series to limit current and measure potential difference. This combination ensures minimal current through the meter while allowing voltage measurement.
See lessAnswer: (D) high resistance in series.
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https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
A current carrying wire kept in a uniform magnetic field will experience a maximum force when it is
The force on a current-carrying wire in a magnetic field is given by: F = BILsinθ The force is maximum when sinθ = 1, i.e., = 90° . This means the wire must be perpendicular to the magnetic field to experience the maximum force. Answer: (A) perpendicular to the magnetic field. For more visit here: hRead more
The force on a current-carrying wire in a magnetic field is given by:
F = BILsinθ
The force is maximum when sinθ = 1, i.e., = 90°
. This means the wire must be perpendicular to the magnetic field to experience the maximum force.
Answer: (A) perpendicular to the magnetic field.
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A straight conducting rod of length l and mass is suspended in a horizontal plane by a pair of flexible strings in a magnetic field of magnitude B. To remove the tension in the supporting strings, the magnitude of current in the wire is :
To remove tension in the strings, the magnetic force must balance the gravitational force: Bil = mg Rearranging for current I: I = mg/BL Thus, the magnitude of current needed to make the net force zero is: I = mg/BL Answer: (C) mg/IB For more visit here: https://www.tiwariacademy.com/ncert-solutionsRead more
To remove tension in the strings, the magnetic force must balance the gravitational force:
Bil = mg
Rearranging for current I:
I = mg/BL
Thus, the magnitude of current needed to make the net force zero is:
I = mg/BL
Answer: (C) mg/IB
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/