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A rectangular coil, of sides 2 cm and 3 ...

A rectangular coil, of sides 2 cm and 3 cm respectively, has 10 turns in it. It carries a current of 1 A, and is placed in a uniform magnetic field of 0.2 T in such a manner that its plane makes an angle `60^(@)` with the field direction. The torque on the loop is

A

`6.0xx10^(-4)Nm`

B

`6.0xx10^(-5)Nm`

C

`1.2xx10^(-3)Nm`

D

6.0Nm

Text Solution

Verified by Experts

The correct Answer is:
A
When a current carrying coil is placed in a uniforme magnetic field B, it experience a torque whose magnetic is given by
`tau=MBsintheta=(NIA)Bsintheta`
where ,M= Magnetic moment of the current of carrying coil
A=Area of the coil
N=Number of turns in the coil
I=Current in the coil
`theta`=Angle between normal to the palne of the coil and the direction of the magnetic field
Here,N=10,I=1A,B=0.2T
` A=2cmxx2cm=6cm^(2)=6xx10^(-4)xxsin30^(@)`
`theta=90^(@)-60^(@)=30^(@)`
`:. "Torque"tau=NIBA sintheta=10xx1xx0.2xx6xx10^(-4)xxsin30^(@)`
`=10xx1xx0.2xx6xx10^(-4)xx(1)/(2)=6xx10^(-4)Nm`
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