Question

We move a square loop of wire 2.3 cm on a side into a region of uniform magnetic field of 1.5 T with the plane of the loop perpendicular to the direction of the field. The loop is moving with a constant velocity of 2.1 m/s into the field region starting from a point outside the field. If the loop has a resistance of 3 Ω, find the force on the loop as it just enters the magnetic field region. (Enter the magnitude.)

Answer 1

Force will be
`F=8.33* 10^(-4)N`

Step-by-step explanation:

We have given side of wire d = 2.3 cm = 0.023 m

Magnetic field B = 1.5 T

Velocity v = 2.1 m/sec

Resistance R = 3 ohm

We know that current is given by
`i=(Bvd)/(R)=(1.5* 2.1* 0.023)/(3)=0.02415A`

Power is given by
`P=i^2R=0.02415^2* 3=0.00174watt`

We know that power is also given by
`P=Fv`

So
`0.00174=F* 2.1`

`F=8.33* 10^(-4)N`