A charged particle of mass 0.0040 kg is subjected to a 6.0T magnetic field which

Question

A charged particle of mass 0.0040 kg is subjected to a 6.0T magnetic field which acts at a right angle 90 to its motion If the particle moves in a circle of radius 0.10 m at a speed of 4.0 m/s. what is the magnitude of the charge on the particle?

A wire is formed into a circle with radius 8.00 mm A current flows through the wire and causes a magnetic field of magnitude B at the center of the loop. If the wire is heated and expands by 3.0%, what does the magnitude of the magnetic field become at the center of the loop?

Explanation / Answer

The magnetic force,F is qvB.
The centripetal force if it is moving in a circle is mv^2/r. Since this is coming from the magnetic force, then
mv^2/r = qvB
q = (mv^2/r)/(vB) = (mv)/(rB) = 0.0040*4.0/0.10* 6.0= 0.0242 C

Magnetic field due to a wire=B=mu_noght*I/2pi*r = 4pie-7*I/2pi*8e-3

Now if the wire expanded by 3% i.e r= 3% of 8mm= 3/100*8e-3=0.00024

r=0.00824 m

Magnetic field due to a wire=B'=mu_noght*I/2pi*r = 4pie-7*I/2pi*0.00824

B'/B= 4pie-7*I/2pi*0.00824/ 4pie-7*I/2pi*8e-3 =8e-3/0.00824=0.97

B'=0.97*B

i.e magnetic field is 97% of that of B

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.