deksteenk9

2022-10-03

a) Define magnetic field.

b) A long straight wire carries 7 A current. At a distance 4 mm from the wire, a particle is moving parallel to it. If the charge of the particle is 8 \mu C and moves with 200 ms^{-1}. Find

1b) the magnetic flux density, 4 mm from the wire.

2b) the magnetic force exerted on the particle.

b) A long straight wire carries 7 A current. At a distance 4 mm from the wire, a particle is moving parallel to it. If the charge of the particle is 8 \mu C and moves with 200 ms^{-1}. Find

1b) the magnetic flux density, 4 mm from the wire.

2b) the magnetic force exerted on the particle.

Quinn Alvarez

Beginner2022-10-04Added 13 answers

Given

Current ,I =7 A

Distance ,D = 4 mm

Charge,Q = 8uC

Velocity, V = 200 m/s parallel to current direction.

Part A.

Magnetism/magnetic force is the force due to moving electric charges, and the nearby space to the moving charge particle, within which the magnetic force can be experienced on other particles is defined as magnetic field.

Part B

1b)

Magnetic flux density is given as

$B={u}_{0}\cdot \frac{I}{2\pi R}\phantom{\rule{0ex}{0ex}}B=2\cdot {10}^{-7}\cdot \frac{7}{4\cdot {10}^{-3}}=3.5\cdot {10}^{-4}\text{}Tesla$

2b)Now force is

$F=q\cdot V\cdot B\cdot \mathrm{sin}(A)$

Where A is angle between B and V direction

$A={90}^{\circ}\phantom{\rule{0ex}{0ex}}F=8\cdot {10}^{-6}\cdot 200\cdot 3.5\cdot {10}^{-4}=5600\cdot {10}^{-10}\phantom{\rule{0ex}{0ex}}F=0.56\text{}uN$

Current ,I =7 A

Distance ,D = 4 mm

Charge,Q = 8uC

Velocity, V = 200 m/s parallel to current direction.

Part A.

Magnetism/magnetic force is the force due to moving electric charges, and the nearby space to the moving charge particle, within which the magnetic force can be experienced on other particles is defined as magnetic field.

Part B

1b)

Magnetic flux density is given as

$B={u}_{0}\cdot \frac{I}{2\pi R}\phantom{\rule{0ex}{0ex}}B=2\cdot {10}^{-7}\cdot \frac{7}{4\cdot {10}^{-3}}=3.5\cdot {10}^{-4}\text{}Tesla$

2b)Now force is

$F=q\cdot V\cdot B\cdot \mathrm{sin}(A)$

Where A is angle between B and V direction

$A={90}^{\circ}\phantom{\rule{0ex}{0ex}}F=8\cdot {10}^{-6}\cdot 200\cdot 3.5\cdot {10}^{-4}=5600\cdot {10}^{-10}\phantom{\rule{0ex}{0ex}}F=0.56\text{}uN$

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