Question

Drive an expression for magmatic field at the center of circular coil

Answer 1

Answer:

The magnetic field strength at the center of a circular loop is given by B=μ0I2R(at center of loop), where R is the radius of the loop. RHR-2 gives the direction of the field about the loop.

Explanation:

Answer 2

Explanation:

Consider a circular coil of radius a and carrying current I in the direction shown in Figure. Suppose the loop lies in the plane of paper. It is desired to find the magnetic field at the centre O of the coil. Suppose the entire circular coil is divided into a large number of current elements, each of length dl. According to Biot-Savart law, the magnetic field  dB  at the centre O of the coil due to current element Idl is given by,

 dB=4πr3μoI(dl×r) 

where  r  is the position vector of point O from the current element. The magnitude of  dB  at the centre O is

 dB=4πa3μoIdlasinθ 

∴dB=4πa2μoIdlsinθ 

The direction of dB  is perpendicular to the plane of the coil and is directed inwards. Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the center O can be found by integrating the above equation around the loop i.e.

∴B=⎰dB=⎰4πa2μoIdlsinθ 

For each current element, angle between  dl and  r  is 90°. Also distance of each current element from the center O is a.

∴B=4πa2μoIsin90o⎰dl 

But ⎰dl=2πa=total length of the coil

∴B=

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