Question:
If \(B\) is magnetic induction, \(e\) is charge of electron, \(m\) is mass and \(c\) is speed of light in vacuum, then the physical quantity having dimensions of
\[
\frac{4\pi mc}{Be}
\]
is
If \(B\) is magnetic induction, \(e\) is charge of electron, \(m\) is mass and \(c\) is speed of light in vacuum, then the physical quantity having dimensions of
\[
\frac{4\pi mc}{Be}
\]
is
Show Hint
Magnetic field dimensions are important:
\[
[B]=MT^{-2}I^{-1}
\]
Use dimensional cancellation carefully.
Updated On: Jun 15, 2026
- Energy
- Electric potential
- Length
- Time
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
To determine a physical quantity, dimensional analysis is used.
\[
[M^aL^bT^cI^d]
\]
Step 1: Write dimensions of each quantity.
Mass: \[ [m]=M \] Velocity: \[ [c]=LT^{-1} \] Charge: \[ [e]=IT \] Magnetic field: \[ [B]=MT^{-2}I^{-1} \]
Step 2: Substitute dimensions.
\[ \left[\frac{mc}{Be}\right] = \frac{(M)(LT^{-1})}{(MT^{-2}I^{-1})(IT)} \] \[ = \frac{MLT^{-1}}{MT^{-1}} \] \[ =L \] Thus dimensions correspond to \[ \boxed{\text{Length}} \] Hence correct option is \[ \boxed{(C)} \]
Step 1: Write dimensions of each quantity.
Mass: \[ [m]=M \] Velocity: \[ [c]=LT^{-1} \] Charge: \[ [e]=IT \] Magnetic field: \[ [B]=MT^{-2}I^{-1} \]
Step 2: Substitute dimensions.
\[ \left[\frac{mc}{Be}\right] = \frac{(M)(LT^{-1})}{(MT^{-2}I^{-1})(IT)} \] \[ = \frac{MLT^{-1}}{MT^{-1}} \] \[ =L \] Thus dimensions correspond to \[ \boxed{\text{Length}} \] Hence correct option is \[ \boxed{(C)} \]
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