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Question

A metallic block carrying current I is subjected to a uniform magnetic induction B as shown in the figure. Find the force experience F by the moving charges. Also find the face, whose potential is lowered result of this force. Assume the speed of the charge carriers to be v.


A
evB^k, ABCD
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B
evB^k, EFGH
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C
evB^k, ABCD
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D
evB^k, EFGH
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Solution

The correct option is A evB^k, ABCD
Here, the charge carriers are electrons, so for current flowing along +x-direction, the electron will be moving towards x-axis.

(i.e), v=v^i and B=B^j

Now, magnetic force on the charge carriers is,

FB=q(v×B)

=e[v^i×B^j]

FB=evB^k

This force will push the electrons towards the face ABCD, lowering its potential.

Hence, the face ABCD will be at lower potential.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.
Why this question ?

To check the understanding of Hall effect.




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