The figure gives the magnetization curve for a paramagnetic material- The vertical axis scale is set by a - 0-15- and the horizontal axis scale is set by b - 0-2 T-K- Let -sam be the measured net magnetic moment of a sample of the material and -max be the maximum possible net magnetic moment of that sample-According to Curie-s law- what would be the ratio -sam-max were the sample placed in a uniform magnetic field of magnitude 0-800 T- at a temperature of 2-00 K-

The section explains the terms used in this problem and the connection between M and #956;. The graph in Figure gives a slope of #160; #160; #160; ...

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Standard XII

Physics

Paramagnetism

The figure gi...

Question

The figure gives the magnetization curve for a paramagnetic material. The vertical axis scale is set by $a = 0.15$ , and the horizontal axis scale is set by $b = 0.2$ T/K. Let $μ_{s a m}$ be the measured net magnetic moment of a sample of the material and $μ_{m a x}$ be the maximum possible net magnetic moment of that sample.According to Curie’s law, what would be the ratio $μ_{s a m} / μ_{m a x}$ were the sample placed in a uniform magnetic field of magnitude $0.800$ T, at a temperature of $2.00$ K?

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Solution

The section explains the terms used in this problem and the connection between M and μ. The graph in Figure gives a slope of
$\frac{M / M_{m a x}}{B_{e x t} / T} = \frac{0.15}{0.20 T / K} = 0.75 K / T$
Thus ,we can write
$\frac{μ}{μ_{m a x}} = (0.75 K / T) \frac{0.800 T}{2.00 K} = 0.30$

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The section explains the terms used in this problem and the connection between M and μ. The graph in Figure gives a slope of M/MmaxBext/T=0.150.20T/K=0.75K/TThus ,we can write μμmax=(0.75K/T)0.800T2.00K=0.30

The section explains the terms used in this problem and the connection between M and μ. The graph in Figure gives a slope of
$\frac{M / M_{m a x}}{B_{e x t} / T} = \frac{0.15}{0.20 T / K} = 0.75 K / T$
Thus ,we can write
$\frac{μ}{μ_{m a x}} = (0.75 K / T) \frac{0.800 T}{2.00 K} = 0.30$