Question

Determine the magnetic force on a wire of 3 cm, carrying a current of 10 A and placed inside a solenoid normal to its axis. Inside the solenoid, the magnetic field is given as 0.27 T.

Answer 1

Length of the wire (l) = 3 cm = 0.03 m

Current flowing in the wire (I) = 10 A

Magnetic field (B) = 0.27 T

Angle between the current and magnetic field, θ=90°.

Magnetic force exerted on the wire is given as:

F=BIlsinθ

= 0.27×10×0.03sin90°

= 8.1×10−2N

Hence, the magnetic force on the wire is 8.1×10−2N. The direction of the force can be obtained from Fleming’s left hand rule.

Answer 2

Answer:

Length of wire, l = 3cm = 0.03 m

Current through wire , I = 10A

Magnetic field , B = 0.27T

wire is placed inside a solenoid perpendicular to its axis. so, θ = 90°

now, F = BIL sinθ [ from formula]

= 0.27 × 10 × 0.03 × sin90°

= 0.081 N

Hence, magnetic force on the wire is 0.081 N