Magnetic Effects Of Current Question 410
Question: OABC is current carrying square loop an electron is projected from the centre of loop along its diagonal AC as shown. Unit vector in the direction of initial acceleration will be
Options:
A) $ \hat{k} $
B) $ -( \frac{\hat{i}+\hat{j}}{\sqrt{2}} ) $
C) $ -\hat{k} $
D) $ \frac{\hat{i}+\hat{j}}{\sqrt{2}} $
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Answer:
Correct Answer: D
Solution:
[d] $ \overset{\to }{\mathop{B}},=-c\hat{k} $ , and $ \overset{\to }{\mathop{v}},=v,\cos ,45^{o}\hat{i}-v,\sin ,45^{o}\hat{j} $ $ =\frac{v}{\sqrt{2}}\hat{i}-\frac{v}{\sqrt{2}}\hat{j} $ . Thus, $ \overset{\to }{\mathop{F}},=q(\overset{\to }{\mathop{v}},\times \overset{\to }{\mathop{B}},)=q[ ( \frac{v}{\sqrt{2}}\hat{i}-\frac{v}{\sqrt{2}}\hat{j} )\times (-c\hat{k}) ] $ $ =qcv[ \frac{\hat{i}+\hat{j}}{\sqrt{2}} ] $
$ \therefore ,\overset{\to }{\mathop{a}},=[ \frac{\hat{i}+\hat{j}}{\sqrt{2}} ] $
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