First slide

Applications involving charged particles moving in a magnetic field

Question

A loop of flexible conducting wire of length l lies in magnetic field ‘B’ which is normal to the plane of loop. A current i is passed through the loop. The tension developed in the wire to open up is

Moderate

A

$\frac{π}{2} B i l$
B

$\frac{B i l}{2}$
C

$\frac{B i l}{2 π}$
D

$B i l$

Solution

The upward magnetic force on element “dl” is balanced by tension force on both ends.
$∴ B i (d l) = 2 T \sin (d θ / 2) \Rightarrow B i (r d θ) = 2 T (d θ / 2) \Rightarrow T = B i r = \frac{B i l}{2 π}$ ( $θ$ is small, $\sin θ = θ$ ).

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