Question

6. A 0.5 m length of current-carrying wire kept perpendicular to a magnetic field of
magnitude 200 mT experiences a force of 1.5 mN. What is the current flowing in the
wire?​

Answer 1

Given :

The length of current carrying wire = l = 0.5 meters

The magnitude of magnetic field = B = 200 × $10^{-3}$  Tesla

The Force exerted = F = 1.5 × $10^{-3}$   Newton

To Find :

The magnitude of current flowing in the  wire

Solution :

Force exerted on current carrying wire is given as

Force = current × length of wire × magnetic field

i.e  F = i l B

or,  1.5 × $10^{-3}$  N  = i × 0.5 m × 200 × $10^{-3}$  Tesla

Or,  1.5 × $10^{-3}$  N = 100 × $10^{-3}$ × i

∴    i = $\dfrac{1.5\times 10^{-3}}{100\times 10^{-3}}$  Amp

i.e  i = $\dfrac{1.5}{100}$ Amp

Or, Current = i = 0.015 Amp

Hence, The magnitude of current flowing in the  wire is 0.015 Amp   Answer