Question

A force acting on a conductor of length 5 m carrying
current of 8 A kept perpendicular to the magnetic field
of 1.5 T is​

Answer 1

Answer:

please follow me dear.

mark as brilliant answer.

Explanation:

Let B be uniform magnetic field ,current in the wire of length L be I, then force due to magnetic field is given by

F=ILxB.

L=0.5m

I=1.2 A

B=2T

Therefore,

|F|=(1.2)(0.5)(2)(sin pi/2)=1.2N. In the direction perpendicular to the plane formed by LandB according to right hand screw law for cross product of two vectors.

Answer 2

Answer:

The force on a current carrying conductor of length 5m carrying current 8A in a perpendicular magnetic field of 1.5 T is 60N .

Explanation:

Force on current carrying conductor placed in a magnetic field :

• If a conductor of length, l carrying a current, i is placed in a magnetic field, B, the magnitude of force on the current carrying conductor is given by :
• F = Bil sinθ
• where, θ is the angle between the direction of magnetic field and direction of flow of current.

Given that :

• Length of the conductor, l= 5 m
• Current in the conductor, i= 8 A
• Magnitude of magnetic field, B = 1.5 T
• The direction of magnetic field is perpendicular to direction of current, θ = 90⁰

Solution :

• The force on the conductor is

• F = Bil sinθ = 1.5 x 8 x 5 x sin(90⁰)
• F = 60 N
• Hence, force on the current carrying conductor is 60 N.