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A wire is bent in the form of an equilateral triangle PQR of side $10 cm$ and carries a current of $5.0 A$ . It is placed in a magnetic field $B$ of magnitude $2 T$ directed perpendicularly to the plane of loop. The force on the three sides of the triangle is [[1]] newton.

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Detailed Solution

A wire is bent in the form of an equilateral triangle PQR of side

10 cm

and carries a current of

5.0 A

. It is placed in a magnetic field

B

of magnitude

2 T

directed perpendicularly to the plane of loop. The forces on the three sides of the triangle is

1 N

By Fleming’s left hand rule

F = BIl

Where

F

denotes force, B denotes magnetic field,

I

represents current and

l

represents the length of the conductor.
Here,

I = 5.0 A, B = 2 T, l = 0.1 m

Therefore the force on each sides of the equilateral triangle is,

F = 5 \times 2 \times 0.1

= 1 N

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