Anonymous

# 普通物理 的問題!! 請幫幫我!!

" A rod of length l=15cm and mass m=30g lies on a plane inclined at

37 度 to the horizontal, as shown in the Fig. below.

A current enters and leaves the rod via light flexible wires which we

ignore. For what current ( magnitude and direction ) will the rod be in

equiliburm in a magnetic field B= 0.25 j T. "

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• Anonymous

First of all, we have to find out, before any current flows in the wire, the normal reaction force from the inclined plane is given by:

mg cos θ where θ is the angle of inclination of the plane to the horizontal.

And dividing this normal reaction force into vertical and horizontal components, we can find out that the vertical one is given by:

mg cos θcosθ = mg cos2 θ

Now, a current I is applied through the wire. In order to make a greater normal reaction force from the inclined plane and by the Flemming's left hand rule, the current should flow in the negative z direction.

And the magnetic force due to this current value is BIl in positive x-direction.

Hence, in order to balance this magnetic force, the inclined plane has to give an extra normal reaction force R such that:

R sin θ = BIl

R = BIl/sin θ

Also, this normal reaction can give an extra vertical component of:

R (vertical) = BIl cos θ/sin θ

So, in addition to the vertical component which already existed before application of the current, we have the total vertical component equal to:

mg cos2 θ + BIl cos θ/sin θ

For making it in equilibrium with the rod's weight:

mg cos2 θ + BIl cos θ/sin θ = mg

cos2 θ + BIl cos θ/(mg sin θ) = 1

BIl cos θ/(mg sin θ) = sin2 θ

BIl cos θ = mg sin3 θ

I = mg sin3 θ/(Bl cos θ)

So putting m = 0.03, g = 10, θ= 37, B = 0.25 T and l = 0.15 m, we have:

I = 2.18 A

Source(s): My physics knowledge