A ring in a magnetic field

Electricity and Magnetism Level 3

A conducting ring of radius R = 10 m R=10 m is placed in a homogeneous magnetic field, perpendicular to its plane. A constant current I = 0.1 A I=0.1 A flows through the ring. It is known that the ring breaks if its tension is greater than 5N. What is the maximum magnetic field B m a x B_{max} in Teslas that one can apply without breaking the ring? Neglect the magnetic field created by the current in the ring.


The answer is 5.

David Mattingly by David Mattingly

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1 solution

David Mattingly Staff
May 13, 2014

Let us consider focus on one half of the ring. The magnetic force acting on an element d l dl is given by d F = I d l B dF= I dl B From symmetry it is clear that the total magnetic force will be directed along the y-axis (we choose the y axis to be perpendicular to the diameter) Denoting the tension in the ring by T, we can write 2 T = ∫ d F y = ∫ d F c o s ( α ) = I B ∫ d l cos ⁡ ( α ) = I B ∫ d l x = I B ( 2 R ) . 2 T= \int dF_{y}= \int dF cos(\alpha) = IB \int dl\cos(\alpha)= I B \int {dl}_{x}= I B (2 R). From this equation, we conclude that B m a x = T m a x I R = 5 T e s l a s . B_{max}= \frac{T_{max}}{I R}= 5 Teslas.

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