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प्रश्न
A current of 2 A enters at the corner d of a square frame abcd of side 20 cm and leaves at the opposite corner b. A magnetic field B = 0.1 T exists in the space in a direction perpendicular to the plane of the frame, as shown in the figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.
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उत्तर
Given:
A square frame abcd of side, l = 20 cm
Electric current through the wire, I = 2 A
Magnetic field, B = 0.1 T
The direction of magnetic field is perpendicular to the plane of the frame, coming out of the plane.
As per the question, current enters at the corner d of the square frame and leaves at the opposite corner b.
Angle between the frame and magnetic field, θ = 90˚
Magnetic force,
`vecF = i vecl xx vecB`
For wire da and cb,
`|vecF| = ilBsin theta`
`|vecF| = ilBsin90°`
`= 2/2 xx 20 xx 10^-2xx 0.1`
= 0.02 N
The direction of force can be found using Fleming's left-hand rule.
Thus, the direction of magnetic force is towards the left.
For wires dc and ab,
`|vecF| = ilBsintheta`
∴ `|vecF| = ilBsin90^circ`
`= 2/2 xx 20 xx 10^-2xx0.1`
`= 0.02 N`
The direction of force can be found using Fleming's left-hand rule.
Thus, the direction of magnetic force is downwards.
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