What is the magnitude of the magnetic force per unit length on a wire carrying a current of 8 A and making an angle 30 degree with the direction of a uniform magnetic field of 0.15 T
Answers
Answer:
Explanation:
Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, image019 NCERT Solutions class-12 physics Moving Charges and Magnetism = 30°.
Magnetic force per unit length on the wire is given as:
f= BI sinimage019 NCERT Solutions class-12 physics Moving Charges and Magnetism
= image020 NCERT Solutions class-12 physics Moving Charges and Magnetism
= 0.6 N m-1
Hence, the magnetic force per unit length on the wire is 0.6 N m-1.
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Answer: 0.6 N /m
Given,
Current on wire , I = 8 A
Angle made by the wire = 30°
Magnetic field , B on the wire = 0.15
We know,
The magnitude of the force on the current carrying conductor is given by ,
F = ILB Sin∅
Where we know the angle is the angle between the direction of the magnetic field and the direction of flow of current.
Thus , Force per unit length is ,
= I B sin ∅
On putting the values ,
= 8 × 0.15 × sin 30°
= 1.2 × 1/2
=0.6
Thus the answer will be 0.6