A thin bar magnet of length 2L is bent at the mid-point so that the angle between them is 60° . The new length of the magnet is:

Question

A thin bar magnet of length 2L is bent at the mid-point so that the angle between them is 60° . The new length of the magnet is: (A)  √2L  (B)  √3L  (C)  2L  (D) L

Tardigrade Admin · Accepted Answer

On bending the magnet, the length of the magnet AC=AB+BC =L sin ( (θ /2) )+L sin ( (θ /2) ) =L sin ( (θ /2) ) =2L sin 30° =2L× (1/2)=L

Q. A thin bar magnet of length $2 L$ is bent at the mid-point so that the angle between them is $60^{\circ}$ . The new length of the magnet is:

$2 L$

$3 L$

$2 L$

$L$

On bending the magnet, the length of the magnet
$A C = A B + BC$
$= L sin (\frac{θ}{2}) + L sin (\frac{θ}{2})$
$= L sin (\frac{θ}{2})$
$= 2 L sin 30^{\circ}$
$= 2 L \times \frac{1}{2} = L$

Q. A thin bar magnet of length 2L is bent at the mid-point so that the angle between them is 60∘ . The new length of the magnet is:

2​L

3​L

2L

L

On bending the magnet, the length of the magnet AC=AB+BC =Lsin(2θ​)+Lsin(2θ​) =Lsin(2θ​) =2Lsin30∘ =2L×21​=L

Q. A thin bar magnet of length $2 L$ is bent at the mid-point so that the angle between them is $60^{\circ}$ . The new length of the magnet is:

$2 L$

$3 L$

$2 L$

$L$

On bending the magnet, the length of the magnet
$A C = A B + BC$
$= L sin (\frac{θ}{2}) + L sin (\frac{θ}{2})$
$= L sin (\frac{θ}{2})$
$= 2 L sin 30^{\circ}$
$= 2 L \times \frac{1}{2} = L$