A horizontal magnetic field of 2.0×10 4 T is at an angle of 33 to the direction

a) we know, magnetic force acting on current carrying conductor, F = B*I*L*sin(theta) = 2*10^-4*12*0.7*sin(33) = 9.15*10^-4 N

ID: 1461213 • Letter: A

Question

A horizontal magnetic field of 2.0×104 T is at an angle of 33 to the direction of the current in a straight, horizontal wire 70 cm long.

If the wire carries a current of 12 A , what is the magnitude of the force on the wire?

Express your answer using two significant figures.

What angle(s) would be required for the force to be half the value found in part A, assuming nothing else is changed?

Express your answer using three significant figures. If there is more than one answer, enter them in ascending order separated by a comma.

we know, magnetic force acting on current carrying conductor,

F = B*I*L*sin(theta)

= 2*10^-4*12*0.7*sin(33)

= 9.15*10^-4 N <<<<<<------------Answer

Apply, F' = B*I*L*sin(theta)

F/2 = B*I*L*sin(theta)

9.15*10^-4/2 = 2*10^-4*12*0.7*sin(theta)

4.575*10^-4 = 16.8*10^-4*sin(theta)

sin(theta) = 4.575/16.8

sin(theta) = 0.2723

theta = sin^-1(0.273)

= 15.8 degrees <<<<<<------------Answer

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