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

Open in App

Solution

Magnetic force per-unit length,
$F / l = B I sin θ = 0.15 \times 8 \times 0.5 = 0.6 N m^{- 1}$

Suggest Corrections

Similar questions

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?

Q. What is the magnitude of magnetic force per unit length on a wirecarrying a current of 8 A and making an angle of 30º with thedirection of a uniform magnetic field of 0.15 T?

Q. The magnetic force per unit length on a wire carrying a current of 10 A and making an angle of

45^{0}

with the direction of a uniform magnetic field of 0.20 T is

Q. A straight wire of length

2 m

carries a current of

10 A

. If this wire is placed in a uniform magnetic field of

0.15 T

making an angle of

45^{\circ}

with the magnetic field, the applied force on the wire will be:

Q. A wire carrying a current of 4 A is in a 0.5 T magnetic field as shown above.
If the direction of electron flow is to the right as shown, what is the magnitude and direction of the force (per unit length) on the wire?

Join BYJU'S Learning Program

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?

Magnetic force per-unit length,F/l=BIsinθ=0.15×8×0.5=0.6 Nm−1

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

Magnetic force per-unit length,
$F / l = B I sin θ = 0.15 \times 8 \times 0.5 = 0.6 N m^{- 1}$