1.

A circular coil of radius 4 cm and of 20 turns carries a current of 3 Amperes. It is placed in a magnetic field at intensity of 0.5Wb//m^(2). What is the magnetic dipole moment of the coil ?

Answer»

`0.15Am^(2)`
`0.3Am^(2)`
`0.45Am^(2)`
`0.6Am^(2)`

SOLUTION :`N=20,I=3A,A=pir^(2)=3.14xx10^(-4)xx16m^(2)`
Magnetic DIPOLE MOMENTS M = NIA
`thereforeM=20xx3xx3.14xx10^(-4)xx16=0.3Am^(2)`


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