First slide

Magnetic force

Question

A metal ring of radius 0.5m with its plane normal to a uniform magnetic field B of induction 0.2T carries a current 100A (clockwise). The tension in newton developed in the ring is

Moderate

A

100
B

50
C

25
D

10

Solution

Given B perpendicular to the plane of the ring, from the figure direction of magnetic force is in positive Y-direction for upper semicircular part.
$M a g n e t i c f o r c e o n u p p e r s e m i c i r c u l a r p a r t, \vec{F} = i (\vec{d l} X \vec{B}) = (i) \times (2 r) \times B \hat{j} = 100 (2×0.5 X \frac{2}{10}) \hat{j} = 20 \hat{j} S o, b a l a n c i n g f o r c e o n s e m i - c i r c u l a r r i n g w e g e t, T e n s i o n i n t w o a r m s = M a g n e t i c f o r c e o n u p p e r h a l f \Rightarrow 2 T = 20 \Rightarrow T = 10 N$

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