a copper strip 20cm wide and 1.5mm thick is placed in a magnectic field of 1.7Wbm . If a current of 220A is set up in the strip, what Hall potential difference appears across the strip? [n=8.5*10^28m^-2, e=1.6*10^19C?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/Hall.html#c2http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/Hall.html#c2
change dimension units to meters. B=1.7
To find the Hall potential difference across the copper strip, we need to use the formula:
V_H = B * I * d / (n * e)
Where:
V_H - Hall potential difference
B - Magnetic field strength
I - Current flowing through the strip
d - Thickness of the strip
n - Charge carrier density
e - Charge of an electron
Given:
B = 1.7 Wbm (tesla)
I = 220 A (amperes)
d = 1.5 mm (millimeters) = 0.0015 m (meters)
n = 8.5 * 10^28 m^-2 (per square meter)
e = 1.6 * 10^19 C (coulombs)
Let's substitute the values into the formula to find the Hall potential difference:
V_H = (1.7 Wbm) * (220 A) * (0.0015 m) / (8.5 * 10^28 m^-2 * 1.6 * 10^19 C)
First, we can simplify the numerator:
V_H = (1.7 * 220 * 0.0015) / (8.5 * 10^28 * 1.6 * 10^19) Wbm * A * m * m / (m^-2 * C)
V_H = 0.5622 / (1.36 * 10^48) Wb * A * m * m / (m^-2 * C)
Next, to simplify the denominator, we multiply 1.36 by 10^48:
V_H = 0.5622 / (1.36 * 10^48) Wb * A * m * m / (m^-2 * C) * (1 / (10^48 * 1.36))
V_H = 0.5622 / 1.36 Wb * A * m * m / (m^-2 * C) * 1 / (10^48 * 1.36)
V_H ≈ 0.4132 * 10^-48 Wb * A * m * m / (m^-2 * C)
Simplifying further, we can cancel out the units to get the final answer:
V_H ≈ 0.4132 * 10^-48 V
Thus, the Hall potential difference across the copper strip is approximately 0.4132 * 10^-48 volts.