Question

A metal wire is bent in the shape of right angle triangle with sides 3 cm, 4 cm, 5 cm is carrying a current of 1 A. The wire loop is in uniform magnetic field 0.5 T whose direction is parallel to the current in the 5 cm side of the loop. The magnitude of magnetic force on the side 3 cm side will b \[\frac{X}{500}N\]. The value of x is ______.

पर्याय

• 3

• 7

• 6

• 9

Answer 1

A metal wire is bent in the shape of right angle triangle with sides 3 cm, 4 cm, 5 cm is carrying a current of 1 A. The wire loop is in uniform magnetic field 0.5 T whose direction is parallel to the current in the 5 cm side of the loop. The magnitude of magnetic force on the side 3 cm side will b \[\frac{X}{500}N\]. The value of x is 6.

Explanation:

The net magnetic field is acting in the direction of GF as shown in figures.

Resolving \[\vec{\mathrm{B}}\] into its components, amongst the components, only \[\overset{\rightarrow}{\operatorname*{\operatorname*{B}}}\sin\theta\] exerts force  in side EF of current carrying loop.

\[\therefore\quad\mathrm{F_{EF}=I\times d(EF)\times B~sin~\theta}\]

From figure (a), \[\mathrm{sin}\theta=\frac{4}{5}\]

\[\therefore\quad\mathrm{F}_{\mathrm{EF}}=1\times0.03\times0.5\times\frac{4}{5}\]

\[\therefore\quad\mathrm{F}_{\mathrm{EF}}=\frac{3}{250}\mathrm{N}=\frac{6}{500}\mathrm{N}\]

\[\therefore\] x = 6