Numerical 1 [2081 B/C]

A wire of length 0.5 m carrying a current of 2A is placed in a uniform magnetic field of strength 0.4T. Find the maximum and minimum force experienced by it.          [2] Ans: 0.4 N and 0 N

Solution

length [latex](l)[/latex] = 0.5 m

Current [latex](I)[/latex] = 2A

Magnetic field strength (B) = 0.4 T

Maximum and minimum force (F) =?

The magnetic force (F) = [latex]BIlsin\theta[/latex]

So, maximum force (F) = [latex]BIl\ when,\ \theta = 90^o[/latex]

Or, F = [latex]0.4\times 2\times 0.5[/latex]

              = 0.4 N

And, minimum force (F) = [latex]BIlsin0^o\ when,\ \theta = 0^o [/latex]

              = 0 N

Numerical 2 [2081 B/C, 2069 ‘A’, 2067]

A slice of indium antimonide 2.5 mm thick carries a current of 150 mA. A magnetic field of flux density 0.5T, correctly applied, produces a maximum Hall voltage of 8.75 mV between the edges of the slice. Calculate the number of free charge carriers per unit volume assuming they each have a charge of [latex]-1.6\times 10^{-19}\ C[/latex]. [2] Ans: [latex]2.14\times 10^{22}\ m^{-3}[/latex]

Solution

Given,

Thickness (t) = 2.5 mm = 2.5 x 10^-3 m

Current (I) = 150 mA = 150 x 10^-3 A

Magnetic flux density (B) = 0.5T

Hall voltage (V_H) = 8.75 mV = 8.75 x 10^-3 V

Concentration of mobile electrons (n) =?

We know that,

Hall voltage (V_H) = [latex]\frac{BI}{net}[/latex]

Or, n = [latex]\frac{BI}{V_Het}[/latex]

              = [latex]\frac{0.5\times (150\times 10^{-3})}{(8.75\times 10^{-3})\times (1.6\times 10^{-19})\times (2.5\times 10^{-13})}[/latex]

              = 2.143 x 10²² electrons/m³.