A circular coil of 25 turns and radius 6.0 cm carrying a current of 10 A, is suspended vertically in a uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil. Calculate the force and torque on coil due to the magnetic field
Question
A circular coil of 25 turns and radius 6.0 cm carrying a current of 10 A, is suspended vertically in a uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil. Calculate the force and torque on coil due to the magnetic field
Solution
To solve this problem, we need to use the formula for the magnetic force on a current-carrying loop, which is given by F = n * I * B * A * sinθ, where:
- n is the number of turns in the coil,
- I is the current,
- B is the magnetic field strength,
- A is the area of the coil, and
- θ is the angle between the normal to the plane of the coil and the magnetic field.
Given that the field lines run horizontally in the plane of the coil, the angle θ is 90 degrees. The sine of 90 degrees is 1, so sinθ = 1.
The area of the coil can be calculated using the formula for the area of a circle, A = πr², where r is the radius of the coil.
Step 1: Calculate the area of the coil A = πr² = π * (0.06 m)² = 0.0113 m²
Step 2: Calculate the force on the coil F = n * I * B * A * sinθ = 25 turns * 10 A * 1.2 T * 0.0113 m² * 1 = 3.39 N
So, the force on the coil due to the magnetic field is 3.39 N.
The torque τ on a current loop in a magnetic field is given by τ = n * I * B * A * sinθ. But since sinθ = 1, the torque is equal to the force calculated above.
So, the torque on the coil due to the magnetic field is also 3.39 N*m.
Similar Questions
A circular coil of radius 5 × 10-2 m and with 40 turns is carrying a current of 0.25 A. Determine the magnetic field of the circular coil at the center.
A coil of wire is mounted between two magnets. When there is a current in the coil,it rotates.The diagram shows the position of the coil, viewed from one end, when it is at an angleof 20° to the horizontal. The current is into the page at X and out of the page at Y.There are 10 turns on the coil and the current in the coil is 6.9 A.Determine the resultant moment about P of the magnetic forces acting on the coil.length of coil = 5.0 cmwidth of coil = 3.5 cmcurrent = 6.9 Anumber of turns = 10magnetic flux density = 0.07 T(4)...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Moment about P = ......................................................(Total for Question 11 = 4 marks)
A coil of radius 44cm has 7 turns. Find the strength of the magnetic field at its centre, if 5A current flows through the coil. (𝜇𝑜=4𝜋×10−7𝑇𝑚/𝐴μ o =4π×10 −7 Tm/A)10×10−5𝑇10×10 −5 T5×10−7𝑇5×10 −7 T5×10−5𝑇5×10 −5 T10×10−7𝑇10×10 −7 T
A rectangular coil of 100 turns and area 500 x 10-4 m2 carrying 2 A current is placed in auniform magnetic field of 10T. Find the maximum torque applying on the coil
A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.250 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A.(a) Find the force on each side of the loop.magnitude µNdirection (b) Find the magnitude of the torque acting on the loop. N · m
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.