A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is Fthe force on the arm AC is A testbook B C
Question
A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is Fthe force on the arm AC is A testbook B C
Solution
The question seems to be incomplete. However, I can provide a general explanation based on the information given.
In a uniform magnetic field, the magnetic force experienced by a current-carrying wire is given by F = ILB*sin(θ), where:
- I is the current through the wire,
- L is the length of the wire,
- B is the magnetic field strength, and
- θ is the angle between the direction of the current and the direction of the magnetic field.
In this case, the magnetic field is acting along AB, so for arm BC, the angle between the current and the magnetic field is 90 degrees. Therefore, the force on arm BC is F = ILBsin(90) = IL*B.
For arm AC, if the current is also flowing perpendicular to the magnetic field, the angle θ is also 90 degrees. Therefore, the force on arm AC would also be F = ILBsin(90) = IL*B.
So, if the lengths of arms AC and BC are equal (since it's an isosceles triangle), and the current through them is the same, the magnetic force on arm AC would be equal to the force on arm BC.
Please provide the complete question or the options for a more accurate answer.
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