Question

An experimenter's diary reads as follow; a charged particle is projected in a magnetic field of $\left(7.0\hat{i}-3.0\hat{j}\right)\times {10}^{-3}T:$ The acceleration of the particle is found to be $\left(\times \hat{i}+7.0\hat{j}\right)\times {10}^{-6}m/s^2$. Find the value of x.

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

Answer: (C)

3

$\text{Given}$: Magnetic field= $(7.0\hat{i}-3.0\hat{j})\times {10}^{-3}T$

Acceleration= $(x\hat{i}+7.0\hat{j})\times {10}^{-6}\frac{m}{s^2}$

$\text{To find}$: Value of x.

$\text{Solutions}$:

Force on the charge particle is given by F=q(V×B)

So F⊥B

∴ F$\cdot B$= 0

i.e. ma$\cdot B$ =0

a$\cdot B$ =0

∴ $(7.0\hat{i}-3.0\hat{j})\cdot (x\hat{i}+7.0\hat{j})$ =0

∴ 7x-21 =0

∴ x= $\frac{21}{7}$

∴ x= 3

$\text{Hence the correct option is C}$