Numerical 1 [2081 GIE ‘B’]

A horizontal straight wire 5 cm long weighing [latex]1.2\ gm^{-1}[/latex] is placed perpendicular to a uniform horizontal magnetic field of flux density 0.6T. If the resistance of wire is [latex]3.8\Omega m^{-1}[/latex]. Calculate the p.d. that has to be applied between the ends of the wire to make it just self-supporting. [3] Ans: [latex]3.7\times 10^{-3}\ V[/latex]

Solution

Length of straight wire [latex](l)[/latex] = 5 cm = 0.05 m

Mass per unit length [latex](\frac{m}{l}) = 1.2\times gm^{-1} = 1.2\times 10^{-3}\ kgm^{-1}[/latex]

So, mass of wire (m) = [latex]\frac{m}{l}\times l = (1.2\times 10^{-3})\times 0.05 = 6\times 10^{-5}[/latex] kg

Magnetic field of flux density (B) = 0.6 T

Resistance per unit length of wire ([latex]\frac{R}{l}[/latex]) = [latex]3.8\Omega m^{-1}[/latex]

So, resistance (R) = [latex]\frac{R}{l}\times l = 3.8\times 0.05 = 0.19\Omega[/latex]

P.d. applied (V) =?

We know that, for self – supporting,

[latex]BIl = mg[/latex]

Or, [latex]B\frac{V}{R}.l = mg[/latex]

Or, [latex]V = \frac{mgR}{Bl}[/latex]     

Or, [latex]V = \frac{(6\times 10^{-5}\times 9.8\times 0.19)}{0.6\times 0.05}[/latex]

= [latex]3.724\times 10^{-3}[/latex] V