The magnetic field at the common center due to a current-carrying loop is given by: B = μ0I/2R Let the currents in loops P and Q be Ip and = Iq = 9A, with radii Rp and Rq such that Rp/Rq = 2/3. For zero net magnetic field: Ip/Iq = Rq/Rp = 3/2 ⟹ Ip = 6A To cancel Q’s anticlockwise field, P must carRead more
The magnetic field at the common center due to a current-carrying loop is given by:
B = μ0I/2R
Let the currents in loops P and Q be Ip and = Iq = 9A, with radii Rp and Rq
such that Rp/Rq = 2/3. For zero net magnetic field:
Ip/Iq = Rq/Rp = 3/2 ⟹ Ip = 6A To cancel Q’s anticlockwise field,
P must carry 6 A in the clockwise direction.
Answer: (D) 6 A in the clockwise direction.
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.
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.
Answer: (D) high resistance in series.
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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.
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
There are unifonn electric and magnetic fields in a region pointing along X-axis. An α-particle is projected along Y-axis with a velocity v. The shape of the trajectory will be :
The magnetic field at the common center due to a current-carrying loop is given by: B = μ0I/2R Let the currents in loops P and Q be Ip and = Iq = 9A, with radii Rp and Rq such that Rp/Rq = 2/3. For zero net magnetic field: Ip/Iq = Rq/Rp = 3/2 ⟹ Ip = 6A To cancel Q’s anticlockwise field, P must carRead more
The magnetic field at the common center due to a current-carrying loop is given by:
B = μ0I/2R
Let the currents in loops P and Q be Ip and = Iq = 9A, with radii Rp and Rq
such that Rp/Rq = 2/3. For zero net magnetic field:
Ip/Iq = Rq/Rp = 3/2 ⟹ Ip = 6A To cancel Q’s anticlockwise field,
P must carry 6 A in the clockwise direction.
Answer: (D) 6 A in the clockwise direction.
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A 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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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
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/