1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
evB^k, EFGH
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
evB^k, ABCD
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
evB^k, EFGH
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A evB^k, ABCDHere, 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.

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program