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Charra [1.4K]
3 years ago
9

a wire carrying 1.5 A passes through a 48 mT magnetic field the wire is peremndicular to the field and makes a quarter circle tu

rn radius 21 cm in the dield regeion. Find the magnitude and direction of the magnetic force
Physics
1 answer:
professor190 [17]3 years ago
8 0

Answer:

Explanation:

Wire is in the form of an arc of the size of a quarter of a circle. The distance between its two end

=√2 x 21 cm

= 29.7 cm

= 29.7 x 10⁻² m

Force on the arc in the magnetic field will be same as the straight line meeting its two ends.

Force = BIL

= 48 x10⁻³ x  1.5 x 29.7 x 10⁻²

= 2138.4 x 10⁻⁵ N.

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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