Question 5

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

Answer

Current in the wire, I = 8 A

Magnitude of the uniform magnetic field, B = 0.15 T

Angle between the wire and magnetic field, Ω¸ = 30°.

Magnetic force per unit length on the wire is given as:

f= BI sinΩ¸

= 0.15 x 8 x 1 x sin30°

= 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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- NCERT Chapter

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