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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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The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 25%, the voltage sensitivity will
The voltage sensitivity of a galvanometer is given by: Voltage Sensitivity = Current Sensitivity/R If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is: 1.2/1.25 = 0.96 This indicates a 4% decrease in voltage sensitivity. Answer: (D) DecreRead more
The voltage sensitivity of a galvanometer is given by:
Voltage Sensitivity = Current Sensitivity/R
If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is:
1.2/1.25 = 0.96
This indicates a 4% decrease in voltage sensitivity.
Answer: (D) Decrease by 4%.
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A current of 5 A is flowing from east to west In a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due south on the table is :
The magnetic field around a long, straight current-carrying wire is given by: B = μ0I/2πr Substituting values: B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is: (B) 1 × 10⁻⁵ T acting upwarRead more
The magnetic field around a long, straight current-carrying wire is given by:
B = μ0I/2πr
Substituting values:
B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T
Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is:
(B) 1 × 10⁻⁵ T acting upwards.
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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 :
An α-particle projected along the Y-axis with velocity v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along theRead more
An α-particle projected along the Y-axis with velocity
v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along the X-axis, resulting in a helical trajectory with its axis parallel to the X-axis.
Answer: (C) Helical with its axis parallel to X-axis.
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