Now let\'s apply the torque equation to a circular coil of wire in a magnetic fi

Question

Now let's apply the torque equation to a circular coil of wire in a magnetic field. The coil has an average radius of 0.0500 m, has 30 turns, and lies in a horizontal plane. It carries a current of 5.00 A in a counterclockwise sense when viewed from above. The coil is in a uniform magnetic field directed toward the right, with magnitude 1.20 T. Find the magnetic moment and the torque on the coil. Which way does the coil tend to rotate? Figure 20.31 shows our diagram. The area of the coil is A = pi r^2 = pi (0.0500 m)^2 = 7.85 times 10^-3 m^2: the angle phi between the direction of B vector and the axis of the coil (perpendicular to its plane) is 90 degree. The magnetic moment for one turn of the coil is mu = IA (Equation 20.9). Therefore, the total magnetic moment for all 30 turns is mu_total = 30_mu = 30 IA = 30 (5.00 A) (7.85 times 10^-3 m^2) = 1.18 A middot m^2. From Equation 20.8, the torque on each turn of the coil is tau = IAB sin phi = (5.00 A) (7.85 times 10^-3 m^2) (1.20 T) (sin 90 degree) = 0.0471 N middot m, and the total torque on the coil of 30 turns is tau = (30) (0.0471 N middot m) = 1.41 N middot m. Using the right-hand rule on the two sides of the coil, we find that the torque tends to rotate the right side down and the left side up. The torque tends to rotate the coil toward the stable equilibrium orientation, in which the normal to the plane is parallel to B vector. Calculate the torque on the coil when it is placed in a magnetic field along the direction of the axis of the coil. Answer: 0 N middot m.

Explanation / Answer

Practice problem

Torque on coil =IABsin Theta

Where theta is the angle between Band A when perpendicular to the plane of coil and magnetic field wil be in same direction,theta wil be 0 and sin theta wil also be 0 hence torque on coil wil be zero

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