Question

a semicircular ring of radius r carrying current I is placed in a magnetic field of intensity B so that plane o wire is perpendicular to magnetic field. net force acting on ring is

Answer 1

Answer:

Explanation:

Magnetic force on the semicircular wire,

F  

mag

​  

 

​  

=i∫(  

dl

×  

B

)=iB∫  

dl

=iB(2r)=2iBr(−  

r

^

) where r is the radius of the semi circular loop.

Answer 2

Answer:

The net force acting on the ring is 2BIr.

Explanation:

The magnetic force is given as,

$F=I(l\times B)=IlBsin\theta$             (1)

Where,

F=force acting on the ring

l=effective length of the ring

I=current flowing through the ring

For a semicircular ring, the effective length will be equal to its diameter i.e.2r.

As per the question,

θ=90°

By substituting the required values in equation (1) we get;

$F=I\times 2r \times Bsin 90\textdegree=2BIr$

Hence, the net force acting on the ring is 2BIr.