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A wire ABCDEF (with each side of length L) bent as shown in the figure and carrying a current I is placed in a uniform magnetic induction B parallel to the positive y-direction. Find the force experienced by the wire and direction?

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ILB, +ve z-axis

ILB, -ve z-axis

-ILB, +ve z-axis

Zero

answer is A.

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Detailed Solution

This force will act along

z axis

so,

ILB is Z axis .

The expression of force is

F = I (l \times B)

The angle between

FE and B

180^{0}

BA and B is 0

For both of these

L \times B = 0

The angle between

ED and B is 90^{0}

Force on

ED

L \times B = ILB sin 90^{0}

= ILB

The angle between

CB and B is 270^{0}

Force on

CB

L \times B = ILB sin 270^{0}

= - ILB

The force on ED or CB acts perpendicular to both

ED and B

CB or B

towards x-axis. These two factors cancel each other as they act opposite
The angle between

DC and B is 90^{0}

Force on

DC

L \times B = ILB sin 90^{0}

= ILB

this force will act along z axis so, ILB is Z axis.

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A wire ABCDEF (with each side of length L) bent as shown in the figure and carrying a current I is placed in a uniform magnetic induction B parallel to the positive y-direction. Find the force experienced by the wire and direction?

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Detailed Solution

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