Physics, asked by warner420, 1 year ago

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

Answered by ROUNAK108
26

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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Answered by Anonymous
79

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  \frac{F}{L} 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 Nm^{-1}

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