Question

# Derive an expression for magnetic induction at a point on the equitorial line of a bar magnet.

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Solution

## Consider a bar magnet NS of length 2l and pole strength m. Let, P be the point on the equatorial line at distance d from mid-point of magnet i.e. O as shown in figure 1.Now, the magnetic induction, B1 at point P due to north pole along NP is,B1=μ04πmNP2B1=μ04πm(d2+l2)2 .........(1) (∵NP2=ON2+OP2)Similarly, the magnetic induction, B2 at point P due to south pole along PS is,B2=μ04πmPS2B2=μ04πm(d2+l2)2 .........(2) (∵PS2=OS2+OP2)From equations (1) and (2), we getB1=B2=B(say)Now, the net magnetic induction BP at P due to bar magnet isBP=B1+B2=2B ........................(3)Resolving B1 and B2 into horizontal and vertical components, as shown in figure 2. The vertical components will be canceled and horizontal will be added together. So, we can write BP=2Bcosθ .............from(3)BP=2(μ04πm(d2+l2)2)(l√(d2+l2)) ............(∵cosθ=l√(d2+l2))For a short dipole, l<<d, thereforeBP=μ04π2mld3BP=μ04πMd3 ........................(∵M=2ml)

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