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A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

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प्रश्न

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

संख्यात्मक
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उत्तर

Given: Length of the wire, l = 3 cm = 0.03 m

Current flowing in the wire, I = 10 A

Magnetic field, B = 0.27 T

Angle between the current and magnetic field, θ = 90°

Formula: F = Bl sin θ

= 0.27 × 10 × 0.03 sin 90°

= 0.081

= 8.1 × 10–2 N

Hence, the magnetic force on the wire is 8.1 × 10–2 N. The direction of the force can be obtained from Fleming’s left-hand rule.

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अध्याय 4: Moving Charges and Magnetism - EXERCISES [पृष्ठ १३५]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 4 Moving Charges and Magnetism
EXERCISES | Q 4.6 | पृष्ठ १३५
एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 4 Moving Charges and Magnetism
Exercise | Q 4.6 | पृष्ठ १६९

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