Question

A current of 19.0 mA is maintained in a single circular loop with a circumference of 2.15 m. A magnetic field of 0.705 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?

N · m

A proton travels with a speed of 4.95 ✕ 10⁶ m/s at an angle of 65° with the direction of a magnetic field of magnitude 0.210 T in the positive x-direction.

(a) What is the magnitude of the magnetic force on the proton?
N

(b) What is the proton's acceleration? (Give the magnitude.)
m/s²

Answer 1

Solution 1)

First, lets find

Radius of the circle from the given data
R = C/2π =2.15 m/2π = 0.342 m

Now,
area =S= πR^2 =0.367 m^2
Now, magnitude of torque ,τ = B*I*S =0.705*19x10^-3*0.367 = 0.0049 N.m (Ans)

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