Question

Having trouble answering these. second photo pertains for question 3.

l. A wire carries a, current of 10.0 A in a direction tha makes an angie of 30.0 with the direction of a magnetic field of strength 0.300 T. Find the magnetic force on 5.00-m length of the wirc. A current of 17.0 mA is maintained in a single circular loop with a circumference of 2.00 m. A magnetic ficld of 0.800 T is directed parallel to the piane of the loop. What is the magnitude of the torque excrted by the magnetic 2 ficld on the loop? 3. Figure P19.29a is a diagram of a device called a velocity selector, in which particles of a specific velocity pass through undeflected while those with greater or lesser velocities are deflected either upwards or downwards. An clectric field is directed perpendicular to a magnetic field, producing an electric force and a magnetic force on the charged particle that can be equal in magnitude and opposite in direction (Fig. P19.29b) and hence cancel. show that particles with a speed of u=E/B will pass through the velocity selector undeflccted.

Answer 1

1.

F^vectoe = IL^vector times B^vector F = BIL sin theta F = (0.300 times 10 times 5 times sin 30 degree)N = 7.5 N 2 T^vector = m^vector times B^vector T = mB sin theta T = BIA sin theta sin theta = sin 90 degree = 1 T = (0.800 times 17 times 10^-3 times pi times 2^2)Nm = 0.1170 Nm 3 F_E = F_B(Forces are balanced) qE = qvB v = E/B

$F=BIL \sin \theta$

2.

$\vec \tau =\vec m \times \vec B$

$\tau =mB \sin \theta$

$\tau =BIA\sin \theta$

$\sin \theta =\sin 90^{\circ}=1$

$\tau =(0.800\times 17\times 10^{-3}\times \pi \times 2^2)\: Nm = 0.170\: Nm$

3.

(Forces are balanced)

$qE=qvB$

$v=\frac {E}{B}$