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Moving Charges and Magnetism question

2011 · Q137
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Moving Charges and Magnetism question

2011 · Q137

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in uniform magnetic field acting along AB. If the magnetic force on the arm BC is F,→\overrightarrow {F,}F,​ the force on the arm AC is

AIPMT 2011 Prelims Physics - Moving Charges and Magnetism Question 58 English
  1. A
    −2 F→- \sqrt 2 \,\overrightarrow F−2​F
  2. B
    −-−  F→\,\overrightarrow FF
  3. C
     F→\,\overrightarrow FF
  4. D
    2 F→\sqrt 2 \,\overrightarrow F2​F
View written solutionFree

Correct answer: B

Here, F→BC=F→{\overrightarrow F _{BC}} = \overrightarrow F FBC​=F

∵F→AB=0 \because {\overrightarrow F _{AB}} = 0∵FAB​=0

The net magnetic force on a current carrying closed loop in a uniform magnetic field is zero.

∴F→AB+F→BC+F→AC=0 \therefore {\overrightarrow F _{AB}} + {\overrightarrow F _{BC}} + {\overrightarrow F _{AC}} = 0∴FAB​+FBC​+FAC​=0

⇒F→AC=−F→BC \Rightarrow {\overrightarrow F _{AC}} = - {\overrightarrow F _{BC}}⇒FAC​=−FBC​    (∵\because∵ F→AB=0{\overrightarrow F _{AB}} = 0FAB​=0)

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