Question

1. A 0.874 m long section of cable carrying current to a car starter motor makes an angle of 64.9º with the Earth’s 5.2×10⁻⁵ T field. What is the current when the wire experiences a force of 6.26×10⁻³ N?

2. An electron moving at 4.55×10³ m/s in a 1.34 T magnetic field experiences a magnetic force of 1.77×10⁻¹⁶ N . What angle does the velocity of the electron make with the magnetic field? Be sure to give an answer between 0 and 90 degrees.

3. What is the maximum torque on a 121-turn square loop of wire 17 cm on a side that carries a 54.8 A current in a 1.7 T field?

Answer 1

1)

F = ILB sin

where

F is force, I is current, L is length

I = F / LBsin

I = 6.26e-3 / 0.874 * 5.2e-5 * sin 64.9

I = 152.1 Amps

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2)

F = qvBsin

sin = F / qvB

sin = 1.77e-16 / 1.6e-19 * 4.55e3 * 1.34

sin = 0.18144

= 10.45 degree

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3)

T = NIABsin

Torque is maximum when   = 90 degree

so,

T = NIAB

T = 121 * 54.8 * 0.172 * 1.7

T = 325.8 N.m