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Question
The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction 1.5×10−2 T. A current of 10.0 ampere flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05m is parallel to the lines of induction, then the torque acting on the loop is
The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction 1.5×10−2 T. A current of 10.0 ampere flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05m is parallel to the lines of induction, then the torque acting on the loop is
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Solution
The correct option is D 6×10−4 Nm
Torque τ acting on a current carrying coil of area A placed in a magnetic field of induction B is given by,
τ=NIBAsinθ
where I= current in the coil, θ= angle which the normal the plane of the coil makes with the lines of induction B.
Here, N=1,B=1.5×10−2 T
A=0.05×0.08=40×10−4 m2
I=10.0amp,θ=90o=π/2
τ=(1.5×10−2)(10.0)×(1)(40×10−4sin(π/2)
τ=6×10−4 Nm
Torque τ acting on a current carrying coil of area A placed in a magnetic field of induction B is given by,
τ=NIBAsinθ
where I= current in the coil, θ= angle which the normal the plane of the coil makes with the lines of induction B.
Here, N=1,B=1.5×10−2 T
A=0.05×0.08=40×10−4 m2
I=10.0amp,θ=90o=π/2
τ=(1.5×10−2)(10.0)×(1)(40×10−4sin(π/2)
τ=6×10−4 Nm
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